

I 

•    "       r- 


•  .     : 


LIBRARY 


UNIVERSITY  OF  CALIFORNIA. 


Class 


Strength   of    Materials 


BY 
H.    E.    MURDOCK,   M.  E.,  C.  E. 

MEMBER  OF  THE  SOCIETY  FOR  THE  PROMOTION  OF  ENGINEERING  EDUCATION 

AND  OF  THE  DEPARTMENT  OF  THEORETICAL  AND  APPLIED  MECHANICS 

IN  THE  UNIVERSITY  OF  ILLINOIS 


FIRST  EDITION 
FIRST  THOUSAND 


NEW  YORK 

JOHN    WILEY   &    SONS 

LONDON:  CHAPMAN  &  HALL,  LIMITED 

1911 


COPYRIGHT,  1911, 

BY 
H.    E.    MURDOCK 


Stanbope  ipress 

F.   H.   GILSON     COMPANY 

BOSTON.   U.S.A. 


PREFACE 


IN  preparing  this  book  the  author  has  had  in  mind 
primarily  the  needs  of  his  own  students  in  strength  of 
materials.  He  hopes,  however,  that  it  will  meet  a  real 
want  in  other  colleges  and  technical  schools  also. 

This  book  has  been  written  with  the  aim  of  making 
intelligible  the  fundamental  principles  of  the  strength 
of  materials  without  the  formal  use  of  the  calculus.  The 
works  which  do  not  use  the  ordinary  calculus  treatment 
usually  omit  some  important  parts  such  as  the  deflec- 
tion of  beams,  strength  of  columns,  horizontal  shear, 
combined  stresses,  impact  loads,  etc.  This  book  is 
designed  to  give  a  fairly  complete  course  in  the  subject 
for  students  who  have  not  had  the  calculus,  or  when 
graphical  presentations  are  preferred.  However,  a  sepa- 
rate chapter  giving  the  derivation  of  the  elastic  curve 
of  beams  by  the  calculus  method  has  been  included  for 
those  who  desire  such  treatment. 

Effort  has  been  made  to  present  the  derivation  of  the 
formulas  in  a  clear  and  concise  manner,  in  such  a  way 
as  to  enable  the  student  to  obtain  an  adequate  compre- 
hension of  the  principles  involved.  While  the  aim  is 
to  emphasize  the  elementary  principles  and  to  develop 
independent  reasoning  in  the  student,  the  ground  covered 
is  that  usually  given  in  a  college  course  for  engineering 
students.  Many  illustrative  examples  and  problems  are 
given  for  the  purpose  of  making  clear  the  application  of 
the  theory.  Answers  to  some  of  the  problems  are  given 
in  order  that  the  student  may  occasionally  check  his 

iii 

226183 


IV  PREFACE 

numerical  work.     The  order  of  the  arrangement  is  one 
that  has  given  good  satisfaction. 

In  the  deduction  of  the  shear  formula  it  is  brought  out 
at  the  first  that  the  shearing  stress  is  not  uniformly  dis- 
tributed over  the  sectional  area  of  the  beam  and  that 
the  maximum  stress  is  greater  than  that  obtained  by 
dividing  the  vertical  shear  by  that  area.  A  chapter  on 
graphic  integration  is  included  and  the  graphical  method 
of  determining  the  deflection  of  beams  is  utilized.  The 
graphical  method  appeals  to  the  eye  as  well  as  to  the 
reason,  and  thus  supplies  an  additional  avenue  of  con- 
ception. It  also  shows  to  advantage  the  meaning  of  the 
constants  of  integration.  The  graphical  method  is  also 
much  more  readily  applicable  to  beams  carrying  non- 
uniform  distributed  loads,  and  to  beams  for  which  the 
moment  of  inertia  of  the  cross  section  is  not  constant. 
When  one  set  of  curves  is  drawn  for  a  given  beam  carry- 
ing a  given  system  of  loading,  those  curves  may  be  used 
for  all  similar  beams  with  similar  loading.  In  the  chapter 
on  the  calculus  method  an  attempt  is  made  to  give  the 
physical  conception  of  the  constants  of  integration 
rather  than  to  treat  them  simply  as  mathematical 
symbols. 

As  the  nature  of  the  behavior  of  columns  under  load 
is  very  uncertain,  the  treatment  given  to  columns  is 
largely  empirical.  Emphasis  is  laid  on  the  straight  line 
formula,  although  the  Euler  and  the  Rankine  formulas 
are  also  given. 

The  author  wishes  to  acknowledge  his  indebtedness 
to  the  following  professors  and  instructors  in  the  College 
of  Engineering  of  the  University  of  Illinois:  to  Dr.  N. 
Clifford  Ricker  for  the  interest  shown  in  the  preparation 
of  this  work,  and  the  use  of  tables  and  data  prepared 
by  him;  to  Professor  A.  N.  Talbot  and  Professor  H.  F. 
Moore  for  many  suggestions  as  to  the  form,  arrange- 


PREFACE  v 

ment,  and  subject  matter,  and  much  assistance  in  the 
preparation  of  the  book;  to  Mr.  G.  P.  Boomsliter  for 
his  criticizing  and  checking  the  examples  and  problems; 
to  Mr.  C.  R.  Clarke  and  Mr.  C.  E.  Noerenberg  for  their 
criticisms  and  help  in  preparing  the  manuscript  for  the 
publishers. 

Although  the  work  has  been  carefully  checked,  errors 
may  exist,  and  for  any  intimation  of  these  I  shall  be 
obliged. 

H.  E.  MURDOCK. 


TABLE    OF   CONTENTS 


CHAPTER  I 

MATERIALS   OF  CONSTRUCTION 

ART.  PAGE 

1.  Introduction i 

2.  Mechanical  and  Physical  Properties 2 

3.  Masonry 3 

(a)  Stone  Masonry 3 

(fe)  Brick  Masonry 3 

(c)  Concrete,  Plain  and  Reinforced 4 

4.  Timber 4 

5.  Cast  Iron 4 

6.  Wrought  Iron 5 

7.  Steel 5 

8.  Other  Materials    6 

Example 7 

Problems .  .                                      8 


CHAPTER  II 

DIRECT  STRESSES 

9.   Definitions 9 

10.  Tension 10 

11.  Compression 12 

12.  Shear 13 

13.  Oblique  Shear 13 

14.  Stress-Deformation  Diagrams 15 

15.  Elastic  Limit  and  Yield  Point 16 

16.  The  Modulus  of  Elasticity 17 

17.  Resilience 18 

18.  The  Shearing  Modulus  of  Elasticity 18 

19.  Poisson's  Ratio 19 

20.  Reduction  of  Area 19 

vii 


viii  TABLE  OF  CONTENTS 

ART.  PAGE 

21.  Uses  of  the  Modulus  of  Elasticity 20 

22.  Stresses  Used  in  Design 20 

Problems 22 


CHAPTER  III 
DIRECT  STRESSES  — APPLICATIONS 

23.  Simple  Cases  of  Direct  Stresses 25 

24.  Stresses  in  Thin  Cylinders 25 

25.  Stresses  in  a  Hoop 26 

26.  Stresses  Due  to  Change  in  Temperature 27 

27.  Stresses  in  Thin  Spheres 28 

28.  Thick  Cylinders  under  Interior  Pressure 28 

29.  Cylinders  under  Exterior  Pressure 29 

Examples 30 

Problems 31 

CHAPTER  IV 

RIVETED   JOINTS 

30.  Riveted  Joints 32 

31.  Kinds  of  Riveted  Joints 32 

32.  Methods  of  Failure  of  Riveted  Joints 33 

33.  Computation  of  Unit  Stresses  Developed  in  Riveted  Joints. . .  34 

34.  Single-Riveted  Lap  Joint 35 

35.  Double-Riveted  Lap  Joint 36 

36.  Lap  Joint  with  More  than  Two  Rows  of  Rivets 37 

37.  Butt  Joint 38 

38.  Compression  Loads  for  Riveted  Joints 38 

39.  Efficiency  of  Riveted  Joints 38 

40.  Design  of  Riveted  Joints 39 

Examples 42 

Problems 44 

CHAPTER  V 
BEAMS 

EXTERNAL  FLEXURAL  FORCES 

41.  Definitions 47 

42.  Methods  of  Loading  Beams / . . . .  48 

43.  Forces  Acting  on  a  Beam  as  a  Whole 49 


TABLE  OF  CONTENTS  IX 

ART.  PAGE 

44.  Forces  Acting  on  a  Portion  of  a  Beam,  Internal  Stresses 51 

45.  Vertical  Shear 52 

46.  Sign  and  Unit  of  Vertical  Shear 53 

47.  Value  of  the  Vertical  Shear  for  Cantilever  and  Simple  Beams . .  53 

48.  Load  and  Shear  Diagrams 54 

49.  Relation  between  the  Load  and  the  Shear 56 

50.  The  Rate  of  Change  of  the  Vertical  Shear 57 

51.  Relation  between  the  Load  and  Shear  Diagrams 57 

52.  Bending  Moment 58 

53.  Sign  and  Unit  of  Bending  Moment 59 

54.  The  Values  of  the  Bending  Moment  at  the  Section  MN,  Dis- 

tant x  from  the  Left  Support  or  Origin  for  Cantilever  and 

Simple  Beams 60 

55.  Bending  Moment  Diagrams 61 

56.  Relation  between  the  Vertical  Shear  and  the  Bending  Moment  62 

57.  The  Rate  of  Change  of  the  Bending  Moment 63 

58.  The  Maximum  Vertical  Shear  and  Bending  Moment 64 

59.  Load,  Shear,  and  Moment  Diagrams  for  Cantilever  and  Simple 

Beams.     Maximum  Shear  and  Moment 65 

60.  Relative  Strength  of  Cantilever  and  Simple  Beams 73 

61.  Moving  Concentrated  Loads  on  a  Beam 74 

CHAPTER  VI 
BEAMS 

INTERNAL  FLEXURAL  STRESSES 

62.  Forces  and  Stresses 83 

63.  Resisting  Shear.     The  Shear  Formula 84 

64.  The  Value  of  k  in  the  Shear  Formula 85 

65.  Resting  Moment 86 

66.  Assumptions  for  the  Resisting  Moment 86 

67.  Distribution  of  the  Fiber  Stresses 87 

68.  Position  of  the  Neutral  Surface  and  the  Neutral  Axis 88 

69.  The  Moment  Formula 90 

70.  Units 91 

71.  Total  Horizontal  Compressive  and  Tensile  Stresses 91 

72.  The  Three  Problems 93 

73.  Modulus  of  Rupture 97 

74.  Maximum  Stress  Diagrams 97 

75.  Beams  of  Uniform  Strength 99 

Examples 99 

Problems 101 


TABLE  OF  CONTENTS 


CHAPTER  VII 

STRESSES  IN   SUCH  STRUCTURES  AS  CHIMNEYS, 
DAMS,  WALLS,  AND  PIERS 

ART.  PAGE 

76.  Kinds  of  Stresses 106 

77.  Eccentric  Loads  on  Short  Prisms 107 

78.  Eccentricity  of  a  Load  that  Will  Produce  Zero  Stress  in  the 

Outside  Fiber 108 

79.  The  Kern no 

80.  Case  of  Eccentric  Loads  Caused  by  a  Combination  of  the 

Weight  of  the  Material  and  Lateral  Pressure no 

81.  Effect  when  the  Line  of  Action  of  the  Resultant  Falls  Outside 

of  the  Kern in 

82.  The  Maximum  Stress  when  the  Line  of  Action  of  the  Resultant 

Falls  Outside  the  Middle  Third  for  Rectangular  Prisms 

which  take  no  Tension 112 

Examples 113 

Problems 114 


CHAPTER  VIII 

GRAPHIC  INTEGRATION 

83.  Definitions 116 

84.  The  First  Method  of  Obtaining  the  Second  Integrated  Curve .  .  117 

85.  The  Second  Method  of  Obtaining  the  Second  Integrated  Curve  1 20 

86.  Constant  of  Integration 124 

87.  Units 126 

Examples 127 

Problems 128 

CHAPTER  IX 
DEFLECTION  OF  BEAMS 

ELASTIC  CURVE 

88.  Bending 130 

89.  The  Radius  of  Curvature  of  Beams 130 

90.  The  Slope  of  the  Neutral  Surface 132 

91.  The  Slope  Curve 134 

92.  The  Rate  of  Change  of  the  Slope 135 

93.  The  Deflection  of  Beams.     The  Elastic  Curve 135 


TABLE  OF   CONTENTS  xi 

ART.  PAGE 

94.  The  Rate  of  Increase  of  the  Deflection 137 

95.  Relations  between  the  Five  Curves 137 

96.  The  Units  for  the  Five  Curves 137 

Example 138 

Problems 139 


CHAPTER  X 
ELASTIC   CURVE 

CANTILEVER  AND  SIMPLE  BEAMS  AND  BEAMS  FIXED  AT  BOTH  ENDS 

97.  Cantilever  Beam,  Concentrated  Load  at  the  End 140 

98.  Cantilever  Beam,  Concentrated  Load  away  from  the  Free  End  144 

99.  Cantilever  Beam,  Uniform  Load 144 

ico.    Cantilever  Beam,  Various  Loading 147 

101.  Simple  Beam,  Concentrated  Load  at  the  Center 148 

102.  Simple  Beam,  Uniform  Load 150 

103.  Beam  Fixed  at  Both  Ends,  Concentrated  Load  at  Center. ...  154 

104.  Points  of  Inflection 157 

105.  Beam  Fixed  at  Both  Ends,  Uniform  Load 158 

106.  Relative  Strength  and  Stiffness  of  Beams 160 

107.  Maximum  Stress  and  Deflection 163 

108.  Relation  between  the  Maximum  Stress  and  the  Maximum 

Deflection 164 

Example 167 

Problems 168 

CHAPTER  XI 
ELASTIC   CURVE 

OVERHANGING,  FIXED  AND  SUPPORTED,  CONTINUOUS  BEAMS 

109.  Overhanging  Beam,  Concentrated  Loads 170 

no.   Overhanging  Beam,  Uniform  Load 172 

in.   Beam  Fixed  and  Supported,  Concentrated  Load  at  Center  172 

112.  Beam,  Both  Ends  Fixed,  Concentrated  Load  at  any  Point.  .  176 

113.  Continuous  Beams 178 

114.  The  Theorem  of  Three  Moments 181 

115.  Hinging  Points  for  Continuous  Beams 185 

Problems 187 


xii  TABLE  OF  CONTENTS 


CHAPTER  XII 

ELASTIC  CURVE  OF  BEAMS  DETERMINED  BY  THE 
ALGEBRAIC  METHOD 

ART.  PAGE 

116.  The  Algebraic  Relations  between  the  Five  Curves 189 

1 1 7.  -The  Choice  of  Coordinate  Axes 192 

118.  The  Constants  of  Integration 192 

119.  Determination  of  the  Constants  of  Integration 193 

Section  of  Zero  Vertical  Shear 195 

Section  of  Zero  Bending  Moment 196 

Section  of  Zero  Slope 196 

Section  of  Zero  Deflection 196 

120.  Essential  Quantities  to  be  Known  about  Beams 196 

Examples 197 

Problems 206 


CHAPTER  XIII 
SECONDARY  STRESSES 

121.  Horizontal  Shear  in  Beams 208 

122.  The  Magnitude  of  the  Horizontal  and  Vertical  Shearing  Unit- 

Stress  at  a  Point 209 

123.  Plate  Girders,  First  Method 212 

124.  Plate  Girders,  Second  Method 212 

125.  Combined  Flexure  and  Tension  or  Compression 214 

126.  Combined    Shearing   Stresses   and  Tensile  or   Compressive 

Stresses 216 

Examples 217 

Problems 219 

CHAPTER  XIV 

COLUMNS  AND  STRUTS 

127.  Discussion 221 

128.  Stiffness  of  Columns 221 

129.  The  Strength  of  Columns 222 

130.  The  Straight-line  Formula 225 

131.  Eccentric  Loads  on  Columns 228 

132.  The  Method  of  Transmitting  Loads  to  Columns 229 


TABLE   OF  CONTENTS  Xlii 

OTHER  COLUMN  FORMULAS 

ART.  PAGE 

133.  Comparative  Strength  and  Stiffness  of  Long,  Ideal  Columns. 

Condition  of  the  Ends 230 

134.  Rankine's  Formula.     Columns  of  Intermediate  Length 231 

135.  Euler's  Formula.     Long  Columns 233 

136.  The  Three  Problems 234 

137.  Eccentric  Loads  on  Columns 234 

138.  Behavior  of  Columns  under  Load 237 

Examples 239 

Problems 243 


CHAPTER  XV 
TORSION 

139.  Stress  and  Deformation.     Round  Shafts 246 

140.  The  Torsion  Formula.     Round  Shafts 247 

141.  Stiffness  of  Shafts 248 

142.  Other  Shapes  of  Cross  Section  of  Shafts 249 

143.  Power  Transmitted  by  Shafts 250 

144.  Combined  Twisting  and  Bending 251 

Examples 251 

.     Problems 252 


CHAPTER  XVI 
REPEATED  STRESSES,  RESILIENCE,  HYSTERESIS  IMPACT 

145.  Repeated  Stresses 255 

146.  Resilience  . 256 

147.  Resilience  of  a  Bar  under  Direct  Stress 258 

148.  Resilience  of  a  Beam 258 

149.  Mechanical  Hysteresis 259 

150.  Lag 260 

151.  The  Effect  of  Rest 260 

152.  Suddenly  Applied  Loads 261 

153.  Impact  Loads 262 

154.  Drop  Loads 264 

Examples 264 

Problems 265 


XIV  TABLE   OF  CONTENTS 


APPENDIX  A 

CENTROIDS  AND  MOMENTS  OF  INERTIA  OF  AREAS 

ART.  PAGE 

Ai 267 

A2.  Centroids  of  Areas 267 

AS.  Axis  of  Symmetry 268 

AI.  Centroid  of  a  Triangle 268 

Ag.  Centroid  of  a  Sector  of  a  Circular  Area 269 

Ae.  Centroids  of  Composite  Areas 271 

AT.  Moment  of  Inertia 272 

Ag.  Radius  of  Gyration 273 

Ag.  Polar  Moment  of  Inertia.  The  Relation  between  the  Polar 

Moment  of  Inertia  and  Ix  and  Iy 273 

AIQ.  Relation  between  Moments  of  Inertia  about  Parallel  Axes  in 

the  Plane  of  the  Area 274 

An.  The  Moment  of  Inertia  of  a  Parallelogram  about  a  Centroidai 

Axis  in  the  Plane  of  the  Area 275 

Ai2.  The  Moment  of  Inertia  of  a  Triangle  about  the  Centroidai 

Axis 279 

AIS.  The  Moment  of  Inertia  of  a  Circular  Area 279 

AH.  Moment  of  Inertia  of  Composite  Areas 281 

Example 282 

Problems 283 

Tables 287 

Index 299 


STRENGTH    OF    MATERIALS 


CHAPTER   I 
MATERIALS  OF  CONSTRUCTION 

i.  INTRODUCTION.  Strength  of  Materials  treats  of 
the  action  of  the  parts  or  members  of  structures  or  ma- 
chines in  resisting  loads  and  other  forces  which  come 
upon  them.  By  the  use  of  the  principles  of  mechanics 
and  the  properties  of  materials,  it  determines  the  internal 
forces  or  stresses  which  are  developed  in  the  simpler 
forms  of  construction,  as  beams  and  columns,  when  they 
are  subjected  to  loads.  The  properties  of  the  engineer- 
ing materials  are  obtained  through  experimental  tests. 
Many  of  the  formulas  derived  in  strength  of  materials 
are  based  on  both  theoretical  analysis  and  experimental 
data,  and  the  subject,  therefore,  is  of  a  semi-empirical 
nature. 

In  architectural  and  engineering  construction,  sta- 
bility, strength,  durability,  and  economy  are  essential 
elements.  The  proper  proportioning,  spacing,  and  con- 
nection of  the  parts  are  important.  Too  little  material 
in  a  member  would  make  the  structure  unsafe,  and  too 
much  would  mean  a  waste.  In  general,  one  member 
should  not  be  designed  in  such  a  way  that  it  will  be 
weaker  than  others  in  the  structure.  Proper  design, 
then,  takes  into  account  the  properties  and  qualities  of 
materials  and  the  mechanics  of  their  action  in  a  struc- 
ture in  such  a  way  as  to  insure  safety  and  economy. 

i 


2  MATERIALS  OF   CONSTRUCTION  [CHAP.  I 

2.  MECHANICAL  AND  PHYSICAL  PROPERTIES.    The 

materials  of  construction  possess  characteristic  proper- 
ties known  as  mechanical  and  physical  properties.  These 
properties  measure  the  fitness  and  ability  of  the  material 
to  sustain  external  loads  or  forces  under  given  con- 
ditions. Different  materials  possess  these  properties  in 
different  degrees,  and,  of  course,  different  grades  of  the 
same  material  differ  in  their  properties.  Some  of  these 
characteristic  properties  can  be  expressed  quantitatively 
between  fairly  well  defined  limits  which  are  determined 
by  test,  while  others  may  be  specified  in  terms  of  ability 
to  withstand  certain  tests  and  fulfill  certain  require- 
ments. The  mechanical  properties  include  strength, 
elasticity,  stiffness,  and  resilience.  Other  physical  prop- 
erties frequently  referred  to  are  toughness,  ductility, 
malleability,  hardness,  fusibility,  and  weldableness. 

When  a  load  is  applied  to  a  piece  or  member  of  a 
structure  the  material  undergoes  a  change  in  size  and 
shape.  If  on  the  removal  of  the  load  the  original  size 
and  shape  are  resumed  the  material  is  said  to  be  elas- 
tic. Elasticity,  then,  is  the  property  of  a  material  by 
which  it  will  regain  its  original  size  and  shape  on  the 
removal  of  an  applied  load.  A  material  which  will  not 
recover  its  original  dimensions  after  deformation  is  termed 
plastic.  If  it  will  only  partially  recover  its  original  di- 
mensions after  deformation  it  is  said  to  be  partially  elastic 
and  partially  plastic.  Most  constructional  materials  are 
nearly  or  quite  perfectly  elastic  up  to  a  certain  limit  of 
deformation,  beyond  which  they  are  partly  elastic  and 
partly  plastic. 

The  ability  to  resist  change  in  shape  and  size  when  a 
load  is  applied  is  termed  stiffness.  In  elastic  materials 
the  amount  of  change  in  size  and  shape  is  generally 
proportional  to  the  amount  of  the  load  applied. 

Materials   will    differ    in    their    tensile,    compressive, 


ART.  3]  MASONRY  3 

and  shearing  strengths.  The  strength  of  a  material  is 
ordinarily  determined  under  the  application  of  a  static 
load  applied  in  a  slowly  increasing  amount.  The  effect 
of  permanent  loads,  of  suddenly  applied  loads,  and  of 
impact  loads,  and  of  the  repetition  of  a  load  many  times, 
requires  separate  consideration. 

A  material  possesses  the  property  of  ductility  if  the 
length  can  be  increased  and  the  cross  section  decreased 
considerably  before  rupture  occurs.  Toughness  is  that 
property  by  which  a  material  will  not  rupture  until 
it  has  deformed  considerably  under  loads  at  or  near 
its  maximum  strength.  This  deformation  may  be  pro- 
duced by  stretching,  bending,  twisting,  etc.  A  tough 
material  gives  warning  of  failure.  It  will  resist  impact 
and  will  permit  rougher  treatment  in  the  manipulations 
which  attend  fabrication  and  use.  A  brittle  material 
will  rupture  without  developing  much  deformation  and 
without  giving  warning.  Brittle  materials  are  unfitted 
to  resist  shock  or  sudden  application  of  load. 

3.  MASONRY.  Masonry  is  mostly  used  to  carry 
compression  loads,  such  as  come  on  foundations,  walls, 
piers,  chimneys,  etc. 

(a)  Stone  Masonry.      The   kinds  of   stone   that   are 
best  adapted  to  building  and  construction  purposes  are 
those  that  can  be  worked  satisfactorily,  can  be  obtained 
in  suitable  size,  have  great  compressive  strength,  and 
are  durable.     Sandstone,  limestone,  marble,  granite,  trap, 
and  slate  are  those  in  most  common  use.     Stone  masonry 
is   laid   up   in   mortar,   and   the   quality  and   character 
specified  will  depend  upon  the  purpose  and  need  of  the 
structure.     The    weight    of    stone    masonry    is    about 
1 60  pounds  per  cubic  foot. 

(b)  Brick  Masonry.     Many  grades  of  brick  are  used. 
This  great  variety  affords  the  designer  opportunity  for 


4  MATERIALS   OF   CONSTRUCTION  [CHAP.  I 

selecting  the  kind  specially  adapted  to  his  purpose. 
Special  kilns  are  required  for  burning  bricks  to  fulfill 
special  requirements,  such  as  paving  brick,  fire  brick, 
pressed  brick,  etc.  The  range  in  the  quality  of  brick 
is  indicated  in  part  by  the  compressive  strength  which 
varies  from  400  pounds  per  square  inch  to  15,000  pounds 
per  square  inch.  The  strength  of  brick  masonry  depends 
largely  upon  the  kind  of  mortar  used  in  the  joints  and 
upon  the  workmanship,  but  it  is  much  smaller  than  that 
of  the  individual  brick,  ranging  from  one-sixth  to  one- 
third  as  much.  The  weight  of  brick  masonry  is  about 
125  pounds  per  cubic  foot. 

(c)  Concrete,  Plain  and  Reinforced.  In  recent  years 
concrete  has  come  into  common  use  for  building  and 
structural  purposes.  The  convenience  with  which  it 
can  be  made  into  the  required  form,  its  durability,  and 
its  fireproofing  qualities  make  it  a  desirable  material. 
For  foundations  and  for  places  where  only  compression 
comes  on  the  structure  the  plain  concrete  is  more  gener- 
ally used,  but  where  tension  exists  steel  is  embedded 
in  the  concrete  to  take  the  tension. 

4.  TIMBER.     Timber  has  been  used  extensively  for 
building  purposes.     There  are  many  varieties  and  qual- 
ities  on    the   market,   affording   good   opportunity   for 
the  selection  of  the  timber  most  suitable  for  the  desired 
purpose.     The  cost  of  timber  is  gradually  increasing, 
and  some  species  have  disappeared  and  others  are  dis- 
appearing   from    the    market    for    structural    purposes. 
The  strength  depends  upon  the  species,  the  condition 
of  growth,  the  seasoning,  the  defects  in  the  timber,  etc. 

5.  CAST  IRON.     Cast   iron   is   a   brittle    metal.     Its 
cheapness,  the  ease  with  which  it  is  cast  into  special 
forms  and  machined  into  exact  shapes,   and  its  high 


ART.  7]  STEEL  $ 

compressive  strength  make  it  valuable  for  a  great  many 
purposes;  but  its  low  tensile  strength,  compared  with 
that  of  other  metals,  and  its  brittleness  make  it  an 
undesirable  material  for  resisting  shock  or  tension. 
Cast  iron  is  made  by  smelting  ore  in  a  blast  furnace. 
In  its  crude  form  it  is  called  pig  iron.  The  strength  of 
cast  iron  and  its  other  properties  vary  widely  and  depend 
upon  the  amount  and  condition  of  the  carbon  and  other 
ingredients  which  it  contains. 

6.  WROUGHT  IRON.     Wrought  iron  is  made  from  pig 
iron  in  a  reverberatory  furnace  by  what  is  called  the 
puddling  process.     The  puddled  balls  obtained   in  the 
process  are  run  through  a  squeezer  and  much  of  the 
cinder  expelled.     The  material  is  rolled  into  muck  bars 
which  are  cut,  piled  together,  heated,  and  finally  rolled 
into  the  shapes  desired.     The  strength  and  other  qual- 
ities depend  upon  the  quality  of  the  pig  iron  used,  and 
upon    the   details    of    manufacture.     Because    wrought 
iron  can  be  easily  worked  and  welded   it  is  adaptable 
to  many  uses,  but  its  use  has  diminished  and  steel  has 
taken  its  place  until  very  little  is  now  manufactured. 

7.  STEEL.     The  term  steel  covers  a  wide  range  of 
material,  —  soft  steel,   mild   steel,   medium   steel,   hard 
steel,  tool  steel,  etc.,  all  being  expressions  used  in  con- 
nection  with   the    various   steels.      In   structural   steel 
the  element  carbon  is  the  one  generally  used  to  control 
strength  and  hardness,  though  other  elements  like  phos- 
phorus, sulphur,  and  manganese  exercise  important  in- 
fluence upon  other  properties. 

The  best  structural  steel  is  made  by  the  open-hearth 
process.  In  this  process  pig  iron,  together  with  scrap 
steel  and  some  iron  ore,  are  melted  in  an  open-hearth 
furnace,  the  carbon,  silicon,  and  other  elements  are 


6  MATERIALS  OF  CONSTRUCTION  [CHAP.  I 

burned  out,  and  a  recarburizer  is  added  to  give  the  proper 
carbon  content  and  to  §  remove  the  iron  oxide  and 
increase  the  manganese,  the  final  product  being  molded 
into  ingots.  Acid  open  hearth  steel  is  produced  in  a 
furnace  which  has  a  siliceous  lining;  no  reagent  is  added 
to  remove  the  phosphorus,  and  hence  the  phosphorus 
content  of  the  product  is  the  same  as  that  of  the  charge. 
Basic  open  hearth  steel  is  produced  in  a  furnace  having 
a  dolomitic  lining  (giving  a  basic  chemical  reaction), 
and  lime  is  added  to  remove  the  phosphorus. 

In  the  Bessemer  process,  melted  pig  iron  is  placed 
in  a  Bessemer  converter  and,  by  the  action  of  air  which 
is  blown  through  the  charge,  most  of  the  carbon,  silicon, 
and  manganese  are  burned  out,  and  a  recarburizer  is 
later  added  to  give  the  proper  carbon  content  and  to 
remove  the  iron  oxide  and  to  increase  the  manganese,  the 
final  product  being  molded  into  ingots.  As  used  in  the 
United  States  the  Bessemer  converters  have  siliceous 
linings,  and  no  phosphorus  is  removed.  Relatively  little 
Bessemer  steel  is  now  used  for  structural  purposes. 

In  the  crucible  process,  crude  wrought  iron  is  fused 
with  a  carbon  flux  in  a  sealed  air-tight  vessel.  The 
crucible  process  is  in  use  for  making  hard  steel,  like 
tool  steel,  spring  steel,  etc. 

The  carbon  content  of  steel  varies  from  less  than  one- 
tenth  per  cent  for  the  softest  steels  to  more  than  one 
and  one-half  per  cent  for  the  hardest  carbon  steels. 
Metals  like  nickel,  tungsten,  vanadium,  etc.,  are  also 
added  to  give  special  amounts  of  strength  or  hardness 
and  produce  grades  of  steel  which  have  special  adapta- 
bility for  various  purposes. 

8.  OTHER  MATERIALS.  Many  other  materials  used 
by  the  engineer  and  architect  are  specially  adapted  to 
the  purpose  for  which  they  are  intended.  Rope  is  made 


ART.  8] 


OTHER   MATERIALS 


of  fibrous  materials  such  as  hemp,  manilla,  cotton,  etc., 
and  of  wire.  Belting  is  made  of  leather,  canvas  and 
rubber,  and  of  metallic  links.  Several  alloys  having 
copper  as  a  basic  element  are  made,  such  as  phosphor 
bronze,  brass,  etc.  Several  kinds  of  artificial  stone  are 
manufactured,  for  most  of  which  sand  and  hydraulic 
cement  are  used  as  the  basic  constituents.  Metals  such 
as  lead  and  aluminum  are  also  used  for  various  purposes. 

Table  I  gives  average  values  of  the  weights  of  various 
materials  used  in  constructional  work,  but  variation 
from  the  tabulated  values  is  to  be  expected. 

TABLE   i 
WEIGHTS  OF  VARIOUS  MATERIALS  USED  IN  CONSTRUCTION 


Material. 

Weight, 
lb.  per  cu.  ft. 

Material. 

Weight, 
lb.  per  cu.  ft. 

Timber 

2S  to  4X 

Sandstone 

I  ZO 

Cast  iron  

450 

Granite  

1  70 

Wrought  iron  . 

480 

Marble  

1  70 

Steel    . 

4OO 

Slate 

1  7  C 

Brass 

ci  c 

Terra  cotta  facing 

I  IO 

Copper,  Bronze 
Aluminum.  .  .  . 
Brick  
Limestone.  .  .  . 

550 
1  60 
100  to  150 

165 

Terra  cotta,  fireproof- 
ing  
Book  tile  
Concrete  

50 
60 

150 

EXAMPLE 

What  is  the  weight  of  a  solid  stone  masonry  pier  with  uni- 
formly sloping  sides  and  rectangular  section,  4  feet  by  8  feet  at 
the  top  and  8  feet  by  16  feet  at  the  bottom  and  20  feet  high? 

This  example  is  most  easily  worked  by  using  the  prismoidal 

formula  to  obtain  the  volume.     This  is  V  =  -  (A  +  4  B  +  C)  in 

6 

which  V  is  the  volume,  h  is  the  altitude,  A  and  C  are  the  areas  of 
the  two  bases,  and  B  is  the  sectional  area  at  the  middle  point. 


MATERIALS  OF   CONSTRUCTION  [CHAP.  I 

For  the  given  example, 

A  =4X8  =  32  square  feet. 

5  =  6X12=72  square  feet. 

C  =  8  X  1 6  =  128  square  feet. 
Therefore  the  weight  of  the  pier  is 
=  i6o  X  -2S°-  (32  +  4  X  72  +  128)  =  160  X  i493l  =  239,000 Ib. 


PROBLEMS 

1.  What  is  the  weight  of  a  wrought  iron  rod  of  i  square  inch 
sectional  area  i  yard  long?  Ans.  10  Ib. 

2.  What  is  the  weight  of  a  hollow  log  3^  feet  in  external  diam- 
eter, 2  feet  in  internal  diameter,  and  16  feet  long? 

3.  What  is  the  weight  per  lineal  foot  of  a  concrete  dam  4  feet 
high,  i  foot  thick  at  the  top,  and  2  feet  thick  at  the  base? 

4.  What  is  the  weight  of  a  solid  granite  obelisk  40  feet  high, 
i  foot  square  at  the  top,  and  3  feet  square  at  the  base? 

Ans.  29,500  Ib. 

5.  What  is  the  weight  of  a  square  brick  chimney  30  feet  high, 
the  inside  dimensions  being  2  feet  at  the  top  and  2\  feet  at  the 
bottom,  and  the  thickness  of  the  walls  uniformly  8  inches? 

6.  A  certain  white  oak  log  12  feet  long  and  2  feet  in  diameter 
weighed  1885  pounds,  what  was  the  weight  of  a  cubic  foot  of  that 
white  oak? 


CHAPTER   II 
DIRECT  STRESSES 

9.  DEFINITIONS.  Force  is  an  action  of  one  body 
upon  another  which  tends  to  change  its  shape  and  to 
produce  a  change  of  motion  in  the  body.  In  this  book 
the  use  of  the  term  will  generally  be  restricted  to  forces 
which  are  externally  applied  to  the  member. 

Stress  is  an  internal  action  which  is  set  up  between 
the  adjacent  particles  of  a  body  when  forces  or  loads 
are  applied  to  the  body.  It  is  developed  whenever  the 
body  undergoes  a  change  in  shape.  Stress  may  be 
considered  an  internal  force. 

A  unit-stress  is  obtained  by  dividing  the  total  stress 
by  the  area  over  which  it  acts  if  the  stress  is  uniformly 
distributed.  In  the  case  of  this  uniform  distribution 
the  unit-stress  is  the  amount  of  stress  per  unit  of  area 
of  the  sectional  area.  If  the  stress  is  not  uniformly 
distributed,  the  unit-stress,  or  the  intensity  of  stress, 
at  a  point  of  the  sectional  area  is  equal  to  the  amount 
of  stress  that  would  be  developed  upon  a  unit  of  area 
if  the  stress  were  uniform  over  the  area  and  if  its 
intensity  were  the  same  as  that  at  the  point. 

Deformation  is  a  change  in  a  dimension  of  a  specimen. 

Shortening  is  a  decrease  in  the  length  of  a  specimen. 

Elongation  is  an  increase  in  the  length  of  a  speci- 
men. 

Detrusion  is  a  lateral  deformation  in  which  the  par- 
ticles apparently  slip  past  each  other.  It  is  caused  by  a 
shearing  force. 

9 


10  DIRECT   STRESSES  [CHAP.  II 

An  axial  load  is  one  whose  line  of  action  coincides 
with  the  axis  of  the  member.  The  axial  load  may  be 
the  resultant  of  several  loads. 

An  axial  stress  is  one  developed  by  an  axial  load. 

If  a  plane  is  passed  perpendicular  to  the  axis  of  a 
bar,  its  intersection  with  the  bar  is  called  the  cross 
section,  or  the  section,  and  its  area  the  sectional  area. 

10.  TENSION.  When  a  load  tends  to  pull  the  par- 
ticles of  a  material  directly  apart  in  the  direction  of 
the  load  the  material  is  under  tension  and  the  load 
is  a  tension  load.  The  internal  stresses  developed 
are  tensile  stresses.  The  resulting  deformation  is  an 
elongation.  As  long  as  rupture  does  not  occur,  the 
forces  acting  on  all,  or  on  a  part  of  the  specimen,  are  in 
equilibrium. 

By  the  principles  of  theoretical  mechanics  it  is  shown 

that  the  conditions  of  equilibrium  are  that  there  shall 

be  no  resultant  force  and  no  resultant  moment.     These 

conditions  are  expressed  in  three  fundamental  equations 

2Fx  =  o,  (i) 

2  Fy  =  O,  (2) 

2  M  =  o.  (3) 

These  conditions  of  equilibrium  are  essential  for  deter- 
mining the  internal  stresses  produced  by  external  forces. 
For  a  homogeneous  specimen  in  direct  tension,  under 


(a)  (6) 

FIG.  i. 

an  axial  load  the  stress  is  uniform  over  the  entire  sec- 
tional area  A.  Let  Fig.  1(0)  represent  the  member 
carrying  the  tension  load  P.  Imagine  the  member 
cut  as  indicated.  Fig.  i(b)  shows  the  left  portion  of 


ART.  10]  TENSION  II 

the  member  with  the  forces  and  stresses  acting  upon  it 
indicated.*  The  total  resisting  stress  is  jA,  where  / 
is  the  tensile  unit-stress  developed.  The  resisting  stress 
is  treated  as  an  external  force  in  the  free-body  diagram, 
then  by  taking  as  the  J^-axis  the  axis  of  the  piece  or 
member,  the  summation  of  the  ^-forces  gives 

2  Fx  =  fA  -  P  =  o, 


If  the  load  becomes  great  enough  to  cause  rupture 
the  maximum  unit-stress  developed  at  any  time  before 
rupture  is  called  the  ultimate  tensile  strength.  In  some 
materials,  such  as  wrought  iron  and  soft  steel,  the  load 
will  increase  to  a  maximum  value,  then  decrease  before 
rupture  occurs.  The  ultimate  strength  differs  for  differ- 
ent specimens  of  the  same  material,  and  for  purposes  of 
design  the  value  should  be  determined  for  each  material 
used  in  the  structure. 

The  unit-stress  at  the  point  of  rupture  is  called  the 
rupturing  strength.  The  rupturing  strength  is  of  no  prac- 
tical value.  For  brittle  materials  the  rupturing  strength 
and  the  ultimate  strength  are  equal. 

When  a  specimen  is  broken  by  a  tension  load,  its 
final  length  will  be  greater  than  its  original  length. 
The  ratio  of  the  increase  in  length  to  the  original  length 
is  called  the  ultimate  elongation.  For  ductile  materials 
the  length  of  the  specimen  has  an  influence  upon  this 
ratio.  So  for  purposes  of  uniformity  the  ultimate 
deformation  is  usually  obtained  for  specimens  of  stand- 
ard size,  either  two  or  eight  inches  in  gauge  length. 
The  average  values  of  the  ultimate  tensile  strength  and 
of  the  ultimate  elongation  for  specimens  of  eight-  inch 
gauge  length  are  given  in  Table  2. 

*  Fig.  i  (&)  is  called  a  free-body  diagram. 


12 


DIRECT  STRESSES 


[CHAP.  II 


TABLE  2 

ULTIMATE  TENSILE  STRENGTH  AND   ULTIMATE  ELONGATION 
OF  MATERIALS 


Material. 

Ultimate  tensile  strength, 
Ib.  per  sq.  in. 

Ultimate  elongation, 
per  cent. 

Timber  

6,000  to  10,000 

I      (T 

Cast  iron  

2O,OOO 

•2 

Wrought  iron  

50,000 

3O.O 

Structural  steel  

6o,OOO 

25  .0  to  30.0 

Steel  wire 

60  ooo  to  250  ooo 

10  o  to  25  o 

II.  COMPRESSION.  When  a  force  acting  on  a  member 
tends  to  push  the  particles  closer  together  in  the  direc- 
tion of  the  force  the  member  is  in  compression.  The 
stresses  arising  are  compressive  stresses.  For  compres- 
sion there  is  a  shortening.  If  the  load  is  axial  and  is 
applied  in  such  a  manner  that  the  stress  developed  is 

uniformly  distributed  over  a  section  of  the  member,  the 

p 

compressive  unit-stress  developed  is  /  =  -j  •  The  aver- 
age values  of  the  ultimate  compressive  strength  are  given 
in  Table  3.  The  table  does  not  include  values  of  the 
ultimate  compressive  strengths  of  malleable  materials. 
Their  values,  however,  should  not  be  considered  greater 
than  the  ultimate  tensile  strengths. 

TABLE  3 

ULTIMATE  COMPRESSIVE  STRENGTH  OF  MATERIALS 


Material. 

Ultimate  compressive 
strength,  Ib.  per  sq.  in. 

Timber 

7.OOO 

Cast  iron 

9O,OOO 

Brick.  .                         .... 

6,OOO 

Brick  masonry  

i,  t;oo 

Rich  concrete  

2,«COO 

Stone  

10,000 

ART.  13] 


OBLIQUE   SHEAR 


12.  SHEAR.  When  external  forces  tend  to  cause  two 
adjacent  sections  of  a  member  to  slip  past  each  other 
the  member  is  in  shear.  Stresses  resisting  such  forces 
are  shearing  stresses.  When  the  two  shearing  forces 
are  near  together  the  shear  is  considered  as  a  simple 


(a) 


(c) 


FIG.  2. 


stress.     Fig.  2  shows  cases  of  direct  shear.     If  a  force  P 

tends  to  shear  a  specimen  along  an  area  A  the  average 

p 


shearing  unit-stress  is  s 

values  of  the  ultimate  strength  in  shear. 


~j'      Table  4  gives  average 


TABLE  4 

ULTIMATE  SHEARING  STRENGTH  OF  MATERIALS 


Material. 

Ultimate  shearing 
strength,  Ib.  per  sq.  in. 

Timber: 
Along  grain  

400 

Across  grain  

3,000 

Cast  iron           

2O,OOO 

Wrought  iron  

40,000 

Structural  steel  

50,000 

Rivet  steel  

45,OOO 

13.  OBLIQUE  SHEAR.  Shearing  stresses  are  developed 
in  structural  members  which  are  subjected  to  direct 
tension  or  compression.  Let  Fig.  3  represent  a  speci- 
men under  the  compression  force  P,  and  imagine  it 
cut  along  the  plane  AB.  The  two  plane  surfaces  made 


14    N  DIRECT   STRESSES  [CHAP.  II 

by  the  cut  would  slip  past  each  other  under  the  action 
of  the  force.  This  tendency  of  the  sections  to  slip  past 
each  other,  which  gives  rise  to  shearing  stresses,  always 
exists  in  members  under  load.  To  deduce  the  value  of 


0v, 


FIG.  3.  FIG.  4. 

the  shearing  unit-stress  developed  along  an  oblique 
plane  making  an  angle  6  with  the  axis,  let  Fig.  4  repre- 
sent the  free-body  diagram  of  one  end  of  the  specimen. 
The  resisting  stresses  acting  on  the  fibers  of  the  plane 
may  be  resolved  into  their  components  Q  and  N  parallel 
and  normal  to  the  plane  respectively.  Taking  the 
X-axis  along  the  plane  there  results 

2  Fx  =  P  cos  0  -  Q  =  o, 

/.    Q  =  Pcos0. 

The  component  Q  parallel  to  the  plane  is  the  force  that 
keeps  this  end  from  slipping  past  the  other  one,  and 
therefore  Q  is  the  resultant  of  the  shearing  stresses,  which 

act  parallel  to  the  plane.     If  A  is  the  area  of  the  cross 

^ 

section  of  the  specimen,  -: — -  is  the  area  of  the  section 

sin  0 

cut.     The  shearing  unit-stress  then  is 

A        P 

s  =  P  cos  6  -T-  - — -  =  -r  sin  6  cos  6. 
sm0      A 

This  is  the  value  of  the  shearing  unit-stress  along  any 
oblique  plane.  To  find  the  value  of  6  for  the  maximum 
shearing  stress  developed  in  a  specimen  the  relation 
sin2  6  +  cos2  0=1  exists.  It  is  shown  by  algebra  that 
when  the  sum  of  two  variables  is  constant  the  product 
of  those  variables  is  a  maximum  when  they  are  equal;  * 
*  See  "Higher  Algebra,"  by  John  F.  Downey,  page  252. 


ART.  14]        STRESS-DEFORMATION  DIAGRAMS  15 

therefore  for  the  maximum  shearing  unit-stress  sin  6  = 
cos  6  which  is  true  when  6  =  45°  then 
_  P       i         i          P_ 

Sm   —    ~   '    ~/=-  *        T=-   ~  .   ' 

A       V2       V2         2  A 

14.  STRESS-DEFORMATION  DIAGRAMS.    Whenever  a 

load  is  applied  to  a  specimen  of  any  material  there  is 
a  corresponding  deformation.  A  graphical  representa- 
tion showing  the  values  of  the  unit-stress  developed  in 


60000 


or         Unit    Deformation  £  , 


0.3 


Structural 
Steel 


FIG.  5. 

the  specimen  along  one  axis  and  the  corresponding 
values  of  its  unit  deformation  along  the  other  axis  is 
called  a  stress-deformation  diagram.  Fig.  5  is  such  a 
diagram  for  a  specimen  of  soft  steel.  The  unit-stress 

p 
f  =  -r  is  plotted  along  the   vertical   axis  and   the   unit 

g 
deformation  e  =  j  is  plotted  along  the  horizontal  axis. 

i 

In  these  equations  P  is  the  total  load  on  the  specimen, 
A  is  the  cross-sectional  area,  e  is  the  total  deformation 


16  DIRECT  STRESSES  [CHAP.  II 

at  the  load  P,  and  /  is  the  original  length  of  the  speci- 
men. In  the  diagram  shown,  the  stresses  measured 
upward  from  0  are  tension  and  those  measured  down- 
ward are  compression,  and  the  deformations  measured 
to  the  right  are  elongations  and  those  measured  to  the 
left  shortenings. 

15.  ELASTIC  LIMIT  AND  YIELD  POINT.  For  stresses 
between  the  two  points  A  and  A',  Fig.  5,  the  deform- 
ation is  proportional  to  the  stress,  while  for  stresses 
beyond  these  points  the  proportionality  does  not  hold. 
These  points  are  the  elastic  limits,  A  in  tension,  A'  in 
compression.  As  long  as  the  stress  is  between  the  two 
values  corresponding  to  A  and  A',  the  specimen  will 
return  to  its  original  length  upon  the  removal  of  the 
load.  If  the  stress  becomes  greater  than  these  values, 
however,  the  length  after  removing  the  load  will  not 
be  the  same  as  before;  the  difference  or  the  change  in 
length  is  the  permanent  set.  The  elastic  limit  is  the 
point  on  the  stress-deformation  curve  where  the  curve 
departs  from  a  straight  line,  or  the  elastic  limit  is  that 
unit-stress  beyond  which  permanent  set  is  developed. 
At  the  point  B,  Fig.  5,  the  deformation  increases  markedly 
with  but  little  increase  in  the  stress.  That  point  is 
the  yield  point.  The  yield  point  then  may  be  defined 
as  the  unit-stress  at  which  there  is  a  marked  increase 
in  the  deformation  with  but  little  or  no  increase  in  the 
stress.  Table  5  gives  average  values  of  the  elastic 
limits  of  wrought  iron  and  steel  as  commonly  determined 
in  the  laboratory.  The  values  for  tension  and  compres- 
sion are  about  the  same.  Values  for  timber  and  cast 
iron  have  not  been  included  on  account  of  the  uncer- 
tainty in  their  determination,  but  when  used  they  may 
be  taken  to  be  about  one-third  to  two-thirds  of  the 
ultimate  strength. 


ART.  1 6] 


MODULUS    OF   ELASTICITY 


TABLE  5 
ELASTIC  LIMIT  OF  WROUGHT  IRON  AND  STEEL 


Material. 

Elastic  limit, 
Ib.  per  sq.  in. 

Wrought  iron  
Structural  steel  
Hard  steel 

25,000 

35.000 
rn  OOO 

i6.  THE  MODULUS  OF  ELASTICITY.  For  values  of 
the  stress  less  than  the  elastic  limit  the  rate  at  which 
the  unit  deformation  increases 
with  the  increase  in  the  unit- 
stress  is  constant,*  i.e.,  the 
unit  deformation  is  propor- 
tional to  the  unit-stress  (Fig. 
6).  This  is  commonly  called 
Hooke's  Law.  Then  for 
stresses  below  the  elastic  limit 
the  unit-stress  divided  by  the 
unit  deformation  gives  a 
constant.  This  ratio  is  the 
modulus  of  elasticity  or  the 
coefficient  of  elasticity. 
Young's  modulus  is  the 
modulus  of  elasticity  for 
direct  tension  or  compression.  FlGt  6> 

If  £  is  the  modulus  of  elasticity,  /  the  unit-stress  below 
the  elastic  limit,  and  e  the  corresponding  unit  deform- 
ation the  value  of  the  modulus  of  elasticity  is 

P 


Ae 


*  Experiments  indicate  that  the  increase  of  deformation  is  not  abso- 
lutely proportional  to  the  increase  in  stress,  but  for  practical  purposes 
they  may  be  taken  as  varying  directly  with  each  other. 


i8 


DIRECT  STRESSES 


[CHAP.  II 


In  the  formula  for  E,  t  is  an  abstract  number;  conse- 
quently the  unit  for  E  is  the  same  as  that  for  /,  pounds 
per  square  inch,  tons  per  square  foot,  etc.  In  Table  6 
are  given  average  values  of  the  modulus  of  elasticity 
in  tension  and  compression  for  some  materials. 

TABLE  6 

MODULUS  OF  ELASTICITY 


Material. 

Modulus  of  elasticity, 
Ib.  per  sq.  in. 

Timber.  

I,5OO,OOO 

Cast  iron  

15,000,000 

Wrought  iron  

25,000,000 

Steel  

30,000,000 

17.    RESILIENCE.      The    stress-deformation    diagrams 
show  that   a  force  acts   through   a   distance  and    thus 

does  work  on  the  speci- 
men. When  the  load  is 
released  the  specimen 
gives  up  energy  stored  in 
it.  This  energy  a  speci- 
men under  stress  is  ca- 
pable of  giving  up  in 
returning  to  its  original 
dimensions  is  called  re- 
silience. 


18.  THE  SHEARING 
MODULUS  OF  ELASTICITY. 

Under  shearing  forces  a 
specimen  will  undergo  a 
detrusion.  The  unit  de- 


FIG.  7. 


trusion  is  obtained   by  dividing  the  total  detrusion  by 
the   length   over  which   it   occurs.      The  ratio  of    the 


ART.  20] 


REDUCTION  OF  AREA 


unit-stress  developed  in  the  specimen  to  the  unit  de- 
trusion  is  called  the  shearing  modulus  of  elasticity. 
It  is  also  called  the  modulus  of  transverse  elasticity, 
and  the  modulus  of  rigidity.  When  a  specimen  is  sub- 
ject to  shearing  stresses  as  in  Fig.  7  the  unit-stress  can 
be  calculated  and  the  detrusion  measured.  If  E8  is 
the  shearing  modulus  of  elasticity,  es  the  unit  detrusion, 
s  the  shearing  unit-stress, 


In  Table  7   are  given  average   values  of  the  shearing 
modulus  of  elasticity. 


TABLE  7 

SHEARING  MODULUS  OF  ELASTICITY 


Material. 

Shearing  modulus  of 
elasticity,  Ib.  per  sq.  in. 

Timber,  across  grain.  .  .  . 
Cast  iron  

400,000 
6,OOO,OOO 

Wrought  iron  

IO,OOO,OOO 

Steel  

I2,OOO,OOO 

19.  POISSON'S  RATIO.     As  the  length  of  a  specimen 
is  increased  by  a  tension  load   the  lateral  dimensions 
decrease.     For  a  compression  load  the  lateral  dimensions 
increase.     For  stresses  below  the  elastic  limit  the  ratio 
of  the  lateral  unit  deformation  to  the  longitudinal  unit 
deformation    is    called    Poisson's   ratio.     This    ratio    is 
considerably  less  than   I   and  for  most  metals  ranges 
between  J  and  |. 

20.  REDUCTION  OF  AREA.    When  a  specimen  is  rup- 
tured by  a  tension  force  the  final  sectional  area  is  less 
than  the  original  area.     The  ratio  of  the  amount  the 


20  DIRECT   STRESSES  [CHAP.  II 

section  at  rupture  is  decreased  to  the  original  area  is 
called  the  reduction  of  area.  Thus,  if  A  i  is  the  original 
area  and  A2  is  the  final  area  at  rupture,  the  reduction 

f  .     AI  —  A2 

of  area  is  —  -=  --- 


21.  USES  OF  THE  MODULUS  OF  ELASTICITY.    The 

Pi 

maximum   unit  -stress   for   which    the   formula   E  =  -A 

Ae 

may  be  used  in  calculations  is  the  elastic  limit.  For 
stresses  below  the  elastic  limit  E  may  be  calculated 
from  data  observed  in  the  laboratory;  and  the  change 
in  length  of  a  specimen  may  be  calculated  by  the  for- 

Pl 
mula  e  =  -j=,  .     The  modulus  of  elasticity  has  an  impor- 

tant application  also  in  determining  the  deflection  of 
beams  and  the  strength  of  columns. 

22.  STRESSES  USED  IN  DESIGN.    In  making  a  design 
safety  and  economy  must  be  considered.     Experiments 
indicate   that   at   stresses   slightly   beyond    the   elastic 
limit  there  is  a  marked  change  in  the  structure  of  the 
material,    and    therefore   the   working    stresses    should 
not    be    carried    beyond    that    value.     At    least    there 
should    not    be    noticeable    permanent    set.     Working 
stresses  are  the  allowable  stresses  used  for  designing; 
they  should  always  be  well  below  the  elastic  limit.     The 
method  of  fixing  upon    values   for   allowable    working 
stresses  is  by  making  a   set  of    experiments  in   which 
the  elastic  limit  and   ultimate  strength  of  a  number  of 
specimens  are  determined,  and  then  by  taking   a  cer- 
tain per  cent  either  of  the  elastic  limit  or  of  the  ulti- 
mate strength  as  the  working  stress.     A  knowledge  of 
the  behavior  of  the  material  under  stress  is  essential  for 
a  proper  determination  of  working  stresses. 


ART.  22] 


WORKING   STRESSES 


21 


TABLE  8 

SAFE  WORKING  STRESSES  IN  POUNDS  PER  SQUARE  INCH  FOR 
STEADY  LOADS 


Material. 

Tension. 

Shear. 

Compression. 

Bending 
(fiber). 

Perpendic- 
ular to 
grain. 

Parallel 
to  grain. 

Timber: 
Cedar,  white  

800 
600 
1,000 
I,2OO 
I,OOO 
800 
800 
I,OOO 

800 

900 
1,000 
1,000 

100 
100 

240 

IOO 
200 

80 

120 
2OO 
1  60 
1  60 
2OO 
125 
IOO 
IOO 

IOO 

IOO 

2,500 

9.500 

10,000 
(    8,000 
(  10,000 

1  80 
1  80 
300 
300 
340 
1  80 
240 
800 
500 
500 
600 
240 
2OO 
2OO 

2OO 
2OO 

,IOO 
,IOO 
,2OO 
,600 

,300 
,OOO 
,2OO 
,800 
,4OO 
,2OO 

,750 
,40O 
I,OOO 
I,2OO 

I,2OO 
I,2OO 

I,OOO 
I,OOO 
I,2OO 
I,2OO 
I,  IOO 

800 
1,300 
1,  800 

1,200 

I,2OO 

1,400 

I,2OO 
I,OOO 
I,  IOO 

1,000 
1,000 

6,000 

I  2  ,OOO 
l6,OOO 

Cypress  .  . 

Elm     

Fir,  Washington  .... 
Gum  

Hemlock  

Larch  

Maple,  sugar  (hard). 
Maple  (average)  
Oak,  red  

white  

Pine,  longleaf  

loblolly  

shortleaf  

yellow,    (Ark., 
etc.)  

800 
800 

3,000 
12,000 
15,000 

Spruce  

Cast  iron  

I  2  ,OOO 
I  2  ,OOO 
1  2  ,OOO 

18,000  (Bearing) 
no 

250 
350 

Wrought  iron' 

Steel,  structural  

rivet 

Brickwork  (in  lime) 

Brickwork  (in  Portland 
cement) 

Concrete     (Portland 
cement) 

Table  8  is  a  modified  extract  from  a  table  of  allowable 
working  stresses  compiled  by  Ricker  from  building  or- 
dinances. A  few  additions  are  given.  A  few  changes 
also  have  been  made  to  agree  with  recent  building 
ordinances. 

The  factor  of  safety  is  defined  as  the  ratio  of  the  ulti- 
mate strength  to  the  working  stress.  This  value  varies 


22  DIRECT  STRESSES  [CHAP.  II 

for  the  different  materials  and  for  the  kind  of  loading. 
Variable  loads  produce  higher  stresses  than  steady  loads 
of  the  same  magnitude.  Suddenly  applied  loads  and 
shocks  produce  higher  stresses  than  variable  loads  of 
the  same  magnitude.  Therefore  the  factor  of  safety 
for  variable  loads  is  usually  taken  about  one-half  greater 
than  that  for  steady  loads,  and  for  sudden  loads  or  shocks 
it  is  two  or  three  times  that  for  steady  loads.  In  speci- 
fications and  building  ordinances  the  allowable  stresses 
are  usually  given  and  also  the  tests  for  the  materials 
specified.  For  such  cases  a  factor  of  safety  has  been 
considered. 


PROBLEMS 

1.  What  must  be  the  height  of  a  brick  tower  if  the  compressive 
unit-stress  on  the  lowest  brick  is  one-tenth  of  its  ultimate  strength? 

2.  Determine   the  shearing  unit-stress  tending   to  shear   off 
the  head  of  a  i|-inch  wrought  iron  bolt  under  a  tension  of  10,000 
pounds,  if  the  head  is  f  inch  deep. 

3.  A  wrought  iron  plate  \  inch  thick  requires  a  force  of  80,000 
pounds  to  punch  a  round  hole  |  inch  in  diameter  through  it. 
Find  the  ultimate  shearing  strength  of  the  plate. 

4.  What  force  is  required  to  punch  a  i-inch  hole  in  a  £-inch 
structural  steel  plate  ? 

5.  In  a  tension  test  of  a  0.19  per  cent  carbon  steel  specimen 
the  diameter  was  0.5  inch,  and  the  gauge  length  was  1.25  inches. 

Each  scale  division  on  the  extensometer  represented inch. 

125,000 

P  is  the  load  in  pounds  and  e  is  the  reading  on  the  exten- 
someter in  scale  divisions.  The  following  readings  were  made; 
P  »  2000,  e  =  60;  P  =  6000,  e  =  180;  P  =  8000,  e  =  240; 


PROB.]  DIRECT   STRESSES  23 

P  =  8000 -|- ,  e  =  290;  P  =  8100,  e  =  6 10.  The  maximum  load 
was  15,600  pounds,  the  corresponding  length  between  punch  marks 
was  1.59  inches.  The  load  at  rupture  was  12,000  pounds,  and 
the  corresponding  length  was  1.76  inches.  The  diameter  at  the 
fracture  was  0.313  inch. 

(a)  Calculate  the  unit-stress  for  each  load. 

Ans.   30,600  Ib.  per  sq.  in.    40,800  Ib.  per  sq.  in. 

(b)  Calculate  the  unit  elongations  for  each  load. 

Ans.  0.00115,  0.00154. 

(c)  Plot  the  stress-deformation  diagram. 

(d)  What  is  the  elastic  limit  ?       Ans.   40,800  Ib.  per  sq.  in. 

(e)  What  is  the  reduction  of  area?  Ans.   60.7%. 

(f)  What  is  the  modulus  of  elasticity  ? 

Ans.    26,600,000  Ib.  per  sq.  in. 

6.  A  wrought  iron  rod  2  inches  square  and  10  feet  long  length- 
ened 0.02  inch  by  suspending  a  load  from  its  lower  end.     Determine 
the  load.   - 

7.  How  much  will  a  roo-ft.  steel  tape,  |  inch  wide  and  $V  inch 
thick,  stretch  under  a  pull  of  40  pounds  ?  Ans.   0.16  in. 

8.  A  vertical  wooden  bar  50  feet  long  and  6  inches  square 
carries  a  load  of  18,000  pounds  at  its  lower  end.     Find  the  unit- 
stress  at  the  upper  end  and  the  elongation  of  the  bar  due  to  the 
combined  weight  of  bar  and  load. 

9.  Determine    the    elongation  of  a   i-inch  wrought  iron  rod 
10  feet  long,  under  a  tensile  load  of  20,000  pounds. 

10.  How  many  ^-inch  square  rods  of  strong  steel  would  be 
required  for  the  suspension  of  a  platform  loaded  with  15  tons, 
if  the  stretching  of  the  rods  is  limited  to  one-half  their  elon- 
gation at  the  elastic  limit?   Each  rod  carries  equal  shares  of  the 
load. 

11.  What  shearing  load  will  a  rivet  £  inch  in  diameter  safely 
carry  ?    A  rivet  f  inch  in  diameter  ? 

12.  What  should  be  the  depth  of  the  head  of  a  bolt  f  inch  in 
diameter  to  carry  safely  the  shear  ? 

13.  What  must  be  the  bearing  area  to  carry  safely  a  load  of 
20,000  pounds  on  a  Washington  fir  beam  ? 

14.  What  should  be  the  sectional  area  of  a  steel  member  (a  —  x) 


DIRECT   STRESSES 


[CHAP.  II 


of  the  truss  shown  in  Fig.  8  ?    If  (a  —  y)  is  a  short  compression 
longleaf  pine  member,  what  should  be  its  section  ? 

15.  Member  (a  —  y),  Fig.  9,  is  a  steel  rod  i£  inches  in  diameter 
stressed  to  its  safe  working  stress.  What  should  be  the  corre- 
sponding sectional  area  of  the  short  white  cedar  compression 
members  (a  —  x)  ? 


FIG.  8 


FIG.  9. 


16.  Design  a  cast  iron  washer  for  the  member  shown  in  Fig.  10 

17.  What  must  be  the  cross  section  of  specimens  of  the  follow- 
ing materials  in  order  that  the  unit-stress  may  be  about  one-third 
the  elastic  limit  under   a  load   of   30,000  pounds   in   tension? 
(a)  Wrought  iron;  (b)  Steel. 

1 8.  In  a  compression  test  of  a  6-inch  concrete  cube  the  load 
was  107,000  pounds  at  rupture.     What  was  the  shearing  unit- 
stress  along  a  plane  inclined  30°  to  the  axis?     What  was  the 
maximum  shearing  stress  developed  ? 

Ans.    1290  Ib.  per  sq.  in.     1490  Ib.  per  sq.  in. 

19.  What   will   be   the   elongation  at   the  elastic  limit   and 
at   rupture  of  an  8-inch  specimen  of  the  following  materials  ? 
(a)  Wrought  iron;  (b)  Structural  steel;  (c)  Hard  steel. 

20.  What  should  be  the  sectional  area  of  a  steel  rod  if  it  is  to 
take  a  tension  load  of  70,000  pounds  ? 

21.  If  a  cast   iron  specimen   1X2  inches  in  sectional  area 
breaks  under  a  tensile  load  of  42,000  pounds,  what  load  will 
probably  break  a  cast  iron  rod  2  inches  in  diameter? 


CHAPTER   III 
DIRECT   STRESSES  —  APPLICATIONS 

23.  SIMPLE  CASES  OF  DIRECT  STRESSES.    The  sim- 
plest cases  of  direct  stresses  are  such  as  exist  in  eyebars, 
belts,   ropes,   cables,   tension   members   in   trusses,  etc. 
For  such  members  the  unit-stress  developed  is  obtained 
by  dividing  the  load  the  member  carries  by  the  sectional 
area  of  the  member.     There  are  other  cases  for  which 
the   stresses   developed    are    practically   direct    stresses 
although  the  line  of  action  of  the  load  may  not  be  along 
the  axis  of  the  member  that  carries  the  load.     Such 
cases  will  be  considered  in  this  chapter. 

24.  STRESSES  IN  THIN  CYLINDERS.    When  a  thin 

cylinder  is  under  interior  pressure,  as  a  steam  boiler, 
water  pipe,  etc.,  the  forces  tending  to  burst  the  cylinder 
act  normally  to  the  inside  surface,  Fig.  n.  These 
forces  develop  internal  tensile  stresses  in  the  metal  of 
the  cylinder.  In  order  to  determine  the  magnitude  of 
these  stresses,  imagine  the  plane  AB,  Fig.  n,  passed 
perpendicular  to  the  page  and  containing  the  axis  of 
the  cylinder.  The  portion  ABD  is  in  equilibrium  under 
the  forces  and  stresses  acting  upon  it,  and  if  that  half 
of  the  cylinder  were  filled  with  some  solid  substance  the 
interior  forces  acting  upon  it  would  be  normal  to  the 
plane  AB\  the  resisting  stresses  also  would  be  normal 
to  that  plane.  The  internal  stresses  actually  developed 
in  the  cylinder  are  the  same  as  would  be  developed  under 
the  imaginary  condition.  Fig.  12  is  a  free- body  diagram 
of  the  part  ABD.  Let  r  be  the  radius,  t  the  thickness, 

25 


26  DIRECT  STRESSES  —  APPLICATIONS     [CHAP.  Ill 

and  /  the  length  of  cylinder,  Q  the  interior  pressure  per 
unit  of  area,  and  /  the  resisting  unit-stress,  which  is 
approximately  uniform  over  the  resisting  area.  Then 
the  force  tending  to  rupture  the  cylinder  is  2  Qlr,  and  the 


FIG.  ii. 


resisting  stress  on  one  side  is  ///  and  on  both  sides  is 
2  ftl.  Then  for  equilibrium 

2  Qlr  =  2  ftl, 
.'•      Qr=fl. 
This  formula  is  true  for  thin  cylinders. 

25.  STRESSES  IN  A  HOOP.  If  hoops  are  heated  and 
then  shrunk  onto  cylinders,  the  unit-stress  can  be  ob- 
tained by  the  application  of  the  formula  for  the  modulus 

PI      fl 
of  elasticity,  E  =  -r-  =  -•»  if  the  difference  between  the 

Ac         6 

normal  diameter  of  the  hoop  when  cool  and  the  one  to 
which  it  is  shrunk  can  be  obtained.  The  effect  is  the 
same  as  if  the  hoop  is  stretched  from  its  normal  diam- 
eter di  to  the  diameter  d  of  the  hoop  when  shrunk  on  the 
cylinder.  The  difference  in  the  length  of  the  hoop  will 


ART.  26] 


TEMPERATURE    STRESSES 


be  TT  (d  —  di)  and  the  unit  elongation  will  be  TT  (d  —  di) 

d  —  d\       d  —  d\  .  _  . 

-T-  irdi  =  — -j —  or  — -j —  approximately,     ror  steel  tires 

CL  i  CL 

a  common  rule  is  to  make  — -, — -  about Actually 

d  1500 

the  final  diameter  d  of  the  hoop  and  cylinder  will  be 
slightly  less  than  the  original  diameter  of  the  cylinder, 
as  the  metal  of  the  cylinder  will  deform  under  the  pres- 
sure due  to  the  hoop.  The  amount  of  this  deformation 
will  depend  upon  the  ratio  of  the  moduli  of  elasticity  of 
the  materials  of  the  cylinder  and  the  hoop,  and  upon  the 
thickness  of  each. 

26.  STRESSES  DUE  TO  CHANGE  IN  TEMPERATURE. 

When  a  material  is  heated  it  will  expand  if  free,  and 
when  cooled  it  will  contract.  If  /°  is  the  change  in 
temperature  and  c  is  the  coefficient  of  expansion,  or 
change  per  unit  of  length  for  one  degree  rise  or  fall, 
the  change  per  unit  of  length  will  be  e  =  ct°.  If  the 

TABLE  9 

COEFFICIENTS  OF  EXPANSION   PER   DEGREE  FAHR. 


Material. 

Coefficient  of  expansion. 

Masonry  
Cast  iron  
Wrought  iron  
Steel  

.  0000050 
.0000062 
.0000067 
.0000065 

member  is  brought  back  to  its  original  length  by  an 
external  force  the  unit-stress  developed  will  be/  =  eE  = 
ct°E.  If,  instead  of  being  allowed  to  change  in  length 
and  then  being  brought  back  to  its  original  length  when 
a  change  in  temperature  takes  place,  the  specimen  is 
rigidly  held  in  its  original  position,  a  unit-stress  of 
/  =  ct°E  will  be  developed.  The  effect  is  the  same  as  if 


28 


DIRECT   STRESSES  — APPLICATIONS    [CHAP.  Ill 


the  specimen  were  allowed  to  change  in  length  and 
then  were  brought  back  to  its  original  length  by  an 
external  force.  Table  9  gives  the  values  of  the 
coefficient  of  expansion  for  each  degree  of  change  in 
temperature  Fahr. 

27.  STRESSES  IN  THIN  SPHERES.  Internal  pressure 
in  domes  or  other  thin  spheres  tends  to  cause  rupture 
around  a  circumference,  (Fig.  13).  By  using  the  same 
nomenclature  as  for  thin  cylinders,  the  force  tending  to 
push  off  the  dome  is  Qirr2  and  the  stress  resisting  this 
force  is/27iT/, 

.'.     Qr  =  2ft. 

This  formula  also  applies  when  an  interior  pressure  acts 
upon  a  cylinder  head. 


FIG.  13. 


FIG.  14. 


28.  THICK  CYLINDERS  UNDER  INTERIOR  PRESSURE. 
If  the  metal  is  thick  in  comparison  with  the  diameter 
of  the  pipe  the  stresses  developed  are  not  direct  stresses 
and  are  not  uniform  throughout  the  thickness  of  the 
metal  of  the  wall  of  the  cylinder,  and  the  formulas  of 
the  previous  paragraphs  cannot  be  used  for  such  cases, 
(Fig.  14).  Various  expressions  for  the  value  of  the 


ART.  29]      CYLINDERS  —  EXTERIOR  PRESSURE  29 

maximum  stress  developed  have  been  deduced.  For 
the  case  when  there  are  longitudinal  and  transverse 
stresses  due  to  the  interior  pressure  Lame*  gives  the 
formula  for  the  maximum  tensile  stress  developed,  which 
comes  on  the  inner  surface  of  the  pipe,  where  r\  is  the 
internal  radius  and  r2  is  the  external  radius, 

f  =  Q  W  +  n2) 
' 


Calvarino*  gives  tor  the  same  case, 
f= 


Birnie*  gives  the  formula  for  thick  guns  when  no  longi- 
tudinal stress  is  developed, 


Any  of  these  formulas  may  be  used  to  investigate  or 
design  thick  pipes  or  guns. 

29.  CYLINDERS  UNDER  EXTERIOR  PRESSURE.  Recent 
experiments  on  the  collapse  of  tubes  under  exterior 
pressure  indicate  that  for  a  length  of  tube  greater  than 
six  times  its  diameter  the  rupturing  pressure  is  indepen- 
dent of  the  length.  The  formulas  deduced  by  Carman 
and  Carr  in  the  University  of  Illinois  Engineering  Exper- 
iment Station  Bulletin  No.  5  are  as  follows:  for  thin 

tubes  where  -.  is  less  than  0.025, 


And  for  thick  tubes  or  ^  greater  than  0.03, 


Q-K'-C.  (2) 

*  See  "Strength  of  Materials,"  by  A.  Morley. 


30  DIRECT  STRESSES  — APPLICATIONS    [CHAP.  Ill 

In  these  formulas  /  is  the  thickness  of  the  walls,  d  is  the 
outside  diameter,  Q  is  the  external  pressure  in  pounds 
per  square  inch  causing  collapse,  and  K,  K',  and  C  are 
experimental  constants.  Table  10  gives  the  values  of 
these  constants. 


TABLE  10 

VALUES  OP  K,  K'.AND  C  FOR  PIPES  UNDER  EXTERIOR  PRESSURE 


Material. 

K. 

K'. 

C. 

Cold-drawn  seamless  steel  
Brass  

50,200,000 
25,150,000 

95.520 

Q?     ?6< 

2,090 

2  4.74. 

Lap-  welded  steel  

83,270 

I  O2  tj 

EXAMPLES 

1.  What  internal  pressure  will  probably  rupture  a  cast  iroh 
pipe  8  inches  in  diameter  and  £  inch  thick  ? 

@4  =  20,000  X  J, 
Q  =  1250  Ib.  per  sq.  in. 

2.  If  a  steel  rail  of  sectional  area  8.8  square  inches  is  subjected 
to  a  drop  in  temperature  of  50°  Fahr.  and  is  prevented  from 
shortening,  what  is  the  force  exerted  upon  it  if  the  initial  force 
was  zero? 

The  unit-stress  developed  is 

.0000065  X  50  X  30.000,000  =  9750  Ib.  per  sq.  in.  in  tension. 
The  total  force  is  8.8  X  9750  =  85,800  Ib. 


PROB.]  DIRECT  STRESSES  —  APPLICATIONS  31 

PROBLEMS 

1.  What  is  the  maximum  tensile  unit-stress  in  a  pipe  24  inches 
in  diameter,  the  plate  being  f  inch  thick,  and  the  internal  pressure 
80  pounds  per  square  inch?  Ans.   2560  Ib.  per  sq.  in. 

2.  What  internal  pressure  will  rupture  a  1 2-inch  steel  pipe 
|  inch  thick?  Ans.  3750  Ib.  per  sq.  in. 

3.  What  should  be  the  thickness  of  the  lower  plates  of  a  steel 
stand-pipe  20  feet  in  diameter  carrying  a  water  pressure  of  80 
pounds  per  square  inch?    Use  a  unit-stress  of  16,000  pounds  per 
square  inch  and  the  efficiency  of  the  joint  75  per  cent. 

Ans.   About  0.8  in. 

4.  What  stress  is  developed  in  a  spherical  steam  dome  10  inches 
in  diameter,  \  inch  thick,  under  a  steam  pressure  of  120  pounds 
per  square  inch  ? 

5.  What  external  pressure  will  cause  a  2-inch  cold-drawn  steel 
tube  |  inch  thick  to  collapse  ? 

6.  What  internal  pressure  will  burst  a  wrought  iron  cylinder 
of  48  inches  inside  diameter  and  f  inch  thickness? 

7.  Determine  the  thickness  of  a  wrought  iron  steam  pipe  18 
inches  inside  diameter  to  resist  a  pressure  of  200  pounds  per  square 
inch  with  an  allowable  stress  of  6000  pounds  per  square  inch. 

8.  What  should  be   the  minimum   thickness  of  a  cast  iron 
sphere  12  inches  inside  diameter   to  withstand  safely  a  steady 
internal  pressure  of  200  pounds  per  square  inch  ? 

9.  What  internal  pressure  will  burst  a  cast  iron  sphere  24  inches 
inside  diameter  and  ^  inch  thick? 

10.  A  short  wrought  iron  bar  i£  inches  in  sectional  area  has 
its  ends  fixed  immovably  between  two  walls  with  no  stress  when 
the  temperature  is  50°  Fahr.     What  pressure  will  be  exerted  on 
the  walls  when  the  temperature  is  1 20°  Fahr.  ? 

11.  Steel  railroad  rails,  each  30  feet  long,  are  laid  at  a  temper- 
ature of  40°  Fahr.    What  space  must  be  left  between  them  in 
order  that  their  ends  shall  just  meet  when  the  temperature  is 
100°  Fahr.  ?    If  the  rails  had  been  laid  with  their  ends  in  contact, 
what  would  be  the  unit-stress  in  them  at  100°  Fahr.  ? 

12.  A  wrought  iron  tie  rod  20  feet  in  length  and  2  inches  in 
diameter  is  screwed  up  to  a  tension  of  10,000  pounds  in  order  to 
tie  together  two  walls  of  a  building.     Find  the  stress  in  the  rod 
when  the  temperature  falls  30°  Fahr,    Also  when  it  rises  20°  Fahr. 


CHAPTER   IV 
RIVETED    JOINTS 

30.  RIVETED  JOINTS.     In   pipes,  tanks,   and   boilers 
made  of  rolled  plates,  the  plates  are  usually  connected 
by  rivets,  and  stress  is  transmitted  from  one  plate  to 
the  other  through  the  rivets.     Such  joints  may  be  called 
boiler,   tank,    or    pipe    joints.     Connections    of    bridge 
members,  and    connections    between    the  members    of 
roof  trusses,   columns,  beams,  etc.,  are  also  made  by 
means  of  rivets.     Such  joints  may  be  called   structural 
joints.     Wherever    pieces   of   metal    are    connected    by 
rivets   the   design   should   give   the   most   efficient   and 
economical   connection  consistent  with   the   given   con- 
ditions.    A  joint  will  fail  at  its  weakest  part,  and  the 
most  efficient  design  will  have  all  parts  of  the  joint  of 
equal  strength. 

Although  the  actual  stresses  developed  in  a  riveted 
joint  may  be  complex,  the  usual  method  is  to  simplify 
the  calculation  by  assuming  conditions  giving  direct 
tensile,  compressive,  and  shearing  stresses.  As  such 
the  stresses  are  easily  computed.  The  treatment  of 
boiler  joints  and  of  structural  joints  is  essentially  the 
same. 

31.  KINDS  OF  RIVETED  JOINTS.     Riveted  joints  may 
also  be  classified  according  to  the  method  of  connecting 
the  plates  and  the  number  of  rows  of  rivets  used.     In 
Fig.   15  are  shown  two  styles  of  lap  joints;  the  main 
plates  overlap  each  other  and  are  connected  by  the 

32 


ART.  32]  FAILURE  OF  RIVETED   JOINTS 


33 


rivets.  Fig.  15  (a)  and  (b)  represent  a  single  riveted 
lap  joint,  and  (c)  and  (d)  represent  a  lap  joint  with 
two  rows  of  rivets,  with  the  rivets  staggered.  Fig.  18 
shows  two  styles  of  butt  joint;  the  edges  of  the  main 
plate  almost  or  wholly  butt  against  each  other,  and  the 
connection  is  made  through  cover  plates.  F.ig.  18  (b) 
shows  a  butt  joint  with  a  single  cover  plate,  and  Fig.  1 8 


A— 

H 

\iii_- 

J 

\ 

G 

H 

\ 

J 

B 

(a) 

.p 
p«H 


(C) 


(6) 


FIG.  15. 


(c)  shows  one  with  a  double  cover  plate.  There  are  other 
styles  of  riveted  joints,  but  the  general  method  of  treat- 
ment is  the  same  for  all  kinds. 


32.  METHODS  OF  FAILURE  OF  RIVETED  JOINTS.   The 

three  principal  ways  in  which  riveted  joints  may  fail 
are  (i)  (for  tension  loads)  by  tension  in  the  plates  along 
AB,  Fig.  15;  (2)  by  crushing  of  the  rivets  along  CD  or 
EF\  (3)  by  shear  of  the  rivets  along  ED.  Besides  these, 
failure  may  occur,  (4),  by  shearing  of  the  plate  along  GH 
and  //;  (5)  by  bending  of  the  rivets;  (6)  by  bending 
of  the  plates,  thus  allowing  the  rivet  heads  to  shear  off 
or  the  rivets  to  fail  in  tension  ;  (7)  by  failure  of  the  plate 


34 


RIVETED   JOINTS 


[CHAP.  IV 


in  tension  along  KL  and  MN  for  staggered  rivets. 
Failure  by  shear  in  the  plates,  number  (4),  is  avoided 
by  making  the  lap  GH  large  enough  to  insure  safety. 
A  rule  sometimes  followed  is  to  make  the  lap  one  and 
one-half  times  the  diameter  of  the  rivet.  Bending  of 
the  rivets,  and  of  the  plates,  numbers  (5)  and  (6),  may 
occur  in  a  single- rive  ted  lap  joint.  Failure  number  (7) 


(6) 


- 

)( 

( 

- 

- 

- 

—  - 

- 

(d) 


FIG.  16. 


is  avoided  by  making  the  distance  between  the  rows  of 
rivets  large  enough  so  that  KL  +  MN  is  somewhat 
greater  than  QR. 

Fig.  1  6  indicates  the  various  ways  in  which  a  riveted 
joint  may  fail. 

(a)  shows  a  failure  due  to  tensile  stress  in  the  plate. 
shows  a  failure  due  to  bearing  stress  in  the  rivet. 
shows  a  failure  due  to  shearing  stress  in  the  rivet. 
shows  a  failure  due  to  the  shearing  stress  in  the 
plate  between  the  edge  of  the  plate  and  the  rivet 
hole. 
(e)  shows  a  failure  by  tension  along  a  staggered  line. 


(b) 
(c) 
(d) 


33.  COMPUTATION  OF  UNIT-STRESSES  DEVELOPED 

IN  RIVETED  JOINTS.  The  calculation  of  the  unit-  stresses 
developed  in  a  riveted  joint  is  made  by  assuming  that 
the  stress  is  uniformly  distributed  over  the  particular 
area  which  is  in  tension,  compression,  or  shear,  and 


ART.  34]  SINGLE-RIVETED   LAP   JOINT  35 

p 

hence  that  the  unit- stress  is  /  =  -r ,  where  P  is  the  total 

A 

load  coming  on  the  area  A  which  resists  tension,  compres- 
sion, or  shear.  In  order  to  determine  the  tensile  stress 
developed,  the  area  of  the  section  subjected  to  tension 
should  be  obtained.  To  determine  the  compressive  or 
bearing  stress,  the  area  subjected  to  bearing  should  be 
found.  For  bearing  it  is  assumed  that  the  stress  on 
one  rivet  is  uniformly  distributed  over  an  area  equal 
to  that  obtained  by  multiplying  the  thickness  of  the 
plate  by  the  diameter  of  the  rivet.  This  area  is  the 
projection  of  the  rivet  on  the  thickness  of  the  plate. 

If  the  entire  stress  transmitted  by  a  rivet  is  taken  by 
a  section  of  the  rivet  at  one  face  of  the  plate  the  rivet 
is  said  to  be  in  single  shear,  and  the  resisting  area  is 
equal  to  the  cross-sectional  area  of  the  rivet.  If  the 
stress  transmitted  is  taken  by  sections  at  two  faces  of 
a  plate  the  rivet  is  said  to  be  in  double  shear,  and  the 
resisting  area  is  equal  to  twice  the  cross-sectional  area 
of  the  rivet.  The  method  of  calculating  the  stresses 
will  be  given  for  a  few  cases. 

34.  SINGLE-RIVETED  LAP  JOINT.  Let  Fig.  17  repre- 
sent a  single- rive  ted  lap  joint  in  which  the  load  P  is 
to  be  transmitted  from  one  plate  to  the  other  by  n 
rivets  in  one  row.  The  load  must  be  transmitted  by 
tension  through  the  plate  past  the  row  of  rivets.  The 
greatest  tensile  unit-stress  developed  will  come  along 
the  section  AB.  Let  ft  be  the  tensile  unit-stress  de- 
veloped, t  the  thickness  of  the  plate,  dr  the  diameter 
of  the  rivet  hole,  and  b  the  width  of  the  plate.  Then 
the  smallest  area  in  tension  taking  the  load  is  t  (b  —  nd'), 

p 

and  the  unit-stress  is  ft  =  .  ,-,       — JTV 

/  (b  —  nd) 

and  P=ftt(b-nd').  (i) 


RIVETED   JOINTS 


[CHAP.  IV 


The  load  brings  compression  on  the  side  of  the  rivet 
as  shown  in  Fig.  17  (b).  If  d  is  the  diameter  of  the 
rivet,  the  total  area  upon  which  the  load  is  assumed  to 

be  distributed  in  compression  or  bearing  is  ntd,  and  the 

p 
bearing  unit-stress  developed  is  /«.  =  — -? 

and  P  =  fcntd.  (2) 

The  load  tends  to  shear  the  rivets  along  the  plane  CD 
between  the  two  plates.  If  fa  is  the  shearing  unit- 


— D 


(a)  (6)  (c* 

FIG.  17. 

stress,  and  A  the  sectional  area  of  each  rivet,  the  total 
resisting  area  in  shear  is  nA,  and  the  shearing  unit- 

p 
stress  is  fa  =  —j 

and  P=fsnA.  (3) 

By  use  of  the  three  formulas  just  developed  the  unit- 
stresses  existing  in  a  single- rive  ted  lap  joint  under  a 
given  load  can  be  calculated,  or  the  load  a  given  joint 
will  carry  can  be  determined,  or  a  joint  can  be  designed 
to  carry  a  given  load.  If  the  plates  connected  are  of 
different  thickness  the  smaller  value  of  t  should  be  used 
in  the  formulas. 

35.  DOUBLE-RIVETED  LAP  JOINT.  The  equations 
representing  the  relation  between  the  load  transmitted 


ART.  36] 


LAP  JOINT 


37 


and  the  unit-stress  developed  for  a  double- rive  ted  lap 

joint  are  similar  to  those  given  for  the  lap  joint  with  a 

single  row  of  rivets. 

For  this  case  let  n  be  the  number  of  rivets  in  one  row, 

and  HI  the  total  number  of  rivets,  then  the  formulas 

become 

P=ftt(b-nd')9  (i') 

P=fenitd,  (2') 

P  =  f8n1A.  (3') 


(a) 


(b) 


(c) 
FIG.  18. 

36.  LAP  JOINT  WITH  MORE  THAN  Two  Rows  OF 

RIVETS.  If  there  are  more  than  two  rows  of  rivets  the 
assumption  is  generally  made  that  the  load  is  distributed 
evenly  among  the  rivets,  and  the  load  each  rivet  carries 
is  obtained  by  dividing  the  total  load  by  the  number 
of  rivets  carrying  the  load.  With  the  same  nomen- 
clature as  in  the  last  article  the  same  formulas  hold  for 


38  RIVETED  JOINTS  [CHAP.  IV 

this  case  (n  would  usually  be  the  number  of  rivets  in 
the  outside  row). 

37.  BUTT  JOINT.    Fig.  18  (b)  shows  a  butt  joint  with 
a  single  cover  plate,  for  which  kind  of  joint  the  formulas 
of  Art.  34  may  be  applied,  if  n  refers  to  the  number  of 
rivets  on  one  side  of  the  seam.     The  similarity  between 
this  joint  and  a  lap  joint  is  readily  seen  by  considering 
the  cover  plate  with  one  side  of  the  main  plate. 

For  a  butt  joint  with  two  cover  plates,  using  the  same 
nomenclature  as  in  Art.  35,  the  formulas  become 

P  =  ftt(b-nd'),  (i") 

P  =  fcnld,  (2"} 

P  =  fsnl2A.  (3") 

Two  sections  of  each  rivet  are  brought  into  shear. 

38.  COMPRESSION  LOADS  FOR  RIVETED  JOINTS.   For 

the  foregoing  analyses  the  loads  have  been  considered 
in  tension.  If  the  member  is  a  compression  member 
the  formulas  for  shear  and  for  compression  in  the  rivets 
will  remain  unchanged;  the  formula  for  compression  in 
the  rivet  need  not  ordinarily  be  considered. 

39.  EFFICIENCY  OF  RIVETED  JOINTS.    The  efficiency 
of  a  riveted  joint  is  the  ratio  of  the  strength  of  the  joint 
to  the  strength   of  the  unpunched   plate.     In   figuring 
the  strength  working  stresses  are  usually  used  in  the 
formulas  to  determine  the  load  P.     When  a  joint  is  de- 
signed, its  strength  under  tension  in  the  plate  calculated 
by  formulas  (i)  Art.  34,  35,  and  37,  the  bearing  strength 
of  the  rivets  calculated  by  formulas  (2),  and  the  shearing 
strength  of  the  rivets  calculated  by  formulas  (3)  should 
be  obtained  and  the  smallest  value  of  the  load  P  taken 
as  the  strength  of  the  joint;  for  if  a  greater  load  is  put 
on  the  joint  one  of  the  safe  stresses  would  be  exceeded. 


ART.  40] 


DESIGN  OF   RIVETED   JOINTS 


39 


This  value  divided  by  the  working  strength  of  the  un- 
punched  plate  is  the  efficiency. 

In  boiler  lap  joints,  for  the  pitch  p,  which  is  the 
distance  from  center  to  center  of  adjacent  rivets  in 
one  row,  the  strength  of  the  unpunched  plate  is  fttp. 
Therefore,  the  efficiency  for  tension  is 

p-d')  =p-d' 

fttp  _  P  _ 

The  efficiency  for  compression  in  the  rivets  is 
fcntd  =fcnd 

UP      ftp  ' 

And  the  efficiency  for  shear  in  the  rivets  is 


= 


' 


_f.nA 
*'  ~  UP 


The  actual  efficiency  of  the  joint  is  the  smallest  one  of 
the  above  values.  Similar  expressions  for  the  various 
other  types  of  riveted  joints  may  be  deduced. 

Table  II  gives  the  range  of  values  for  efficiency  for 
the  types  of  boiler  joints  listed  for  ordinary  design. 

TABLE  ii 

EFFICIENCY   OF  JOINTS 


Kind  of  joint. 

Efficiency, 
per  cent. 

Single-riveted  lap  joint  

50-65 

Double-riveted  lap  joint  

65-75 

Single-riveted  butt  joint  

65-75 

Double-riveted  butt  joint  

70-80 

Triple- rive  ted  joints  are  frequently  used  and  high  effi- 
ciencies are  obtained. 

40.  DESIGN  OF  RIVETED  JOINTS.  For  use  with  ordi- 
nary thickness  of  plates  in  structural  work  f-inch  and 
f-inch  rivets  are  the  prevailing  sizes.  For  light  work 


40  RIVETED   JOINTS  [CHAP.  IV 

^-inch  and  f-inch  rivets  are  used.  In  specially  heavy 
sections  larger  rivets  are  used,  ij-inch  rivets  being 
occasionally  used. 

The  pitch  of  the  rivets  to  be  used  in  a  design  depends 
upon  the  kind  of  joint  used  and  the  purpose  of  the  joint. 
For  tanks  and  boilers  the  joint  must  be  tight  as  well  as 
strong;  therefore  the  spacing  should  be  small.  For 
structural  members  strength  is  the  main  object  to  be 
accomplished. 

The  diameter  of  the  rivet  hole  is  somewhat  larger  than 
the  diameter  of  the  rivet.  In  boiler  joints  the  diameter 
of  the  rivet  hole  is  usually  assumed  equal  to  or  TV  inch 
larger  than  that  of  the  rivet,  and  in  structural  joints  it 
is  assumed  f  inch  larger. 

The  ideal  joint  for  strength  and  economy  would  be 
the  one  that  would  be  of  equal  strength  in  tension  in 
the  plate,  bearing  in  the  rivets,  and  shear  in  the  rivets. 
For  this  to  be  the  case  the  three  values  of  the  allowable 
resisting  stresses  as  calculated  by  the  three  formulas, 
Art.  34,  35,  and  37,  would  be  the  same;  therefore,  for 
single- riveted  lap  joints, 

ftt(b-nd')  =  fentd,' 
and 

fcntd  =fsnA, 
and 

ftt(b-nd')  =  fsnA. 

Similar  equations  can  be  written  for  the  other  types  of 
riveted  joints.  * 

In  practice  it  is  not  usually  necessary  or  practicable 
to  make  such  ideal  joints,  and  the  resulting  efficiency 
will  be  somewhat  less  than  that  of  the  ideal  joint. 

Limitations  of  the  size  of  rivet,  conditions  for  tight- 
ness of  joints,  convenience  for  shop  work,  and  many 
other  items  may  prevent  making  joints  of  equal  strength 
in  tension,  bearing,  and  shear,  and  small  variations  from 


ART.  40]  DESIGN  OF   RIVETED   JOINTS  41 

the   ideal   conditions   will   not   materially   decrease   the 
efficiency. 

For  the  design  of  boiler,  tank,  or  pipe  joints  the  follow- 
ing procedure  is  a  convenient  one  and  is  recommended. 

(a)  Decide    upon   the   working   stresses   for   tension, 
shear,  and  bearing,  and  calculate  f,  :ft,  a.ndfc:ft. 

(b)  Select  the  type  of  joint  to  suit  the  conditions. 

(c)  In    the    first    calculations  assume    the  efficiency, 
calculate    the   necessary   thickness,    and    then   select   a 
commercial  thickness  of  plate. 

(d)  Determine   the   limiting   size   of   rivet   for   shear 
or    bearing.     The    general    limiting  size    for    any   style 
of  joint  may  well  be  expressed  in  terms  of  the  thick- 
ness  of  the  plate.     For  example,  for  fa'ft  =  2:3  and 
fc'-ft  =  3'-  2    in    the   design   of   riveted    lap   joints   any 
diameter   of   rivet   less   than   2.87  /  will   not   involve  a 
question   of   bearing   strength.     Also   in   ordinary   butt 
joints  having  double  cover  plates,  the  bearing  strength 
of  the  rivet  will  not  need  consideration  if  the  diameter 
of  the  rivet  does  not  exceed  1 .43  t. 

(e)  Select  a  working  size  of  rivet  within  this  limit. 

(f)  With  /  and  d  determined  calculate  the  pitch  by 
equating  the  strength   in  tension  and  the  strength  in 
shear  or  bearing,   using  shear  or  bearing  according  to 
which  controls  the  strength  for  the  type  of  joint  used. 
Only  in  special  or  unusual  types  of  joints  will  tension 
and  shear  govern. 

(g)  Calculate  the  efficiency  of  the  joint  and  the  stresses 
in  the  rivets  and  plate  to  see  that  the  working  stresses 
are  not  exceeded. 

Practice  is  not  uniform  in  regard  to  the  values  of  the 
allowable  unit-stresses  to  be  used  in  design.  The 
following  values  may  be  used  in  solving  problems  in 
this  course:  /,  equals  8000  pounds  per  square  inch, 
ft  equals  12,000  pounds  per  square  inch,  and  fc  equals 


42  RIVETED   JOINTS  [CHAP.  IV 

18,000  pounds  per  square  inch.     The  resulting  ratios 

are  ~  =  ~  =  -  •     The  assumption  made  for  the  bearing 
Jt      Jc      3 

area  of  the  rivets  is  only  approximate  and  experiments 
show  high  values  for  the  ultimate  crushing  strength 
figured  on  that  basis ;  therefore  a  high  value  for  the  crush- 
ing unit-stress  can  be  assumed. 


EXAMPLES 

i.  Select  two  channels  for  the  lower  chord  of  a  truss  in  which 
the  maximum  stress  is  49,300  pounds  in  tension.  Also  determine 
the  number  of  f-inch  rivets,  if  |-inch  gusset  plates  are  used,  for  the 
connection.  Use  ft  =  15,000  pounds  per  square  inch,/c  =  18,000 
pounds  per  square  inch,  and  /«  =  8000  pounds  per  square  inch. 
The  net  sectional  area  required  is 

At  =  49,300-:-  15,000  =  3. 29 square  inches. 
By  use  of  a  hand  book,  the  section  is  found  first. 
Try  two  2-inch  by  2-inch  by  ^-inch  angles. 
The  gross  area  is 

Ag  =  2  X  1.75  =  3.5  square  inches. 

By  counting  the  diameter  of  the  hole  £-inch  larger  than  that  of 
the  rivet,  the  effective  tension  area  will  be  reduced  by  the  amount 

Ai=  2  X  f  X  \  =  .875  square  inch. 
The  effective  area  then  would  be 

Ae  =  Ag  —  Ai  =  3.500  -  .875  =  2.625  square  inches. 
This  is  too  small.     Try  two  3^-inch  by  3-inch  by  A-inch  angles. 
Ag  =  2  X  1.94  =  3.88  square  inches. 
Ai  =  2XjiXr5  =  .547  square  inch. 
The  effective  area  then  is 

Ae  =  Ag  -  Ai  =  3.88  -  .55  =  3.33  square  inches. 
This  is  a  little  in  excess  of  the  required  area;  therefore  use  two 
3 1-inch  by  3-inch  by  A-inch  angles. 

To  determine  the  number  of  rivets  necessary  for  bearing,  the 


EXAM.]  RIVETED  JOINTS  43 


required  bearing  area  is  Ac=        —  =  2.74  square  inches.    The 

18,000 

greatest  bearing  stress  will  be  developed  between  the  gusset  plate 
and  the  rivet,  since  the  thickness  of  the  two  angles  is  greater  than 
that  of  the  gusset  plate.  The  bearing  area  for  one  rivet  is 

dt  =  f  X  f  =  .281  square  inch. 
Therefore,  the  number  of  rivets  required  for  bearing  is 

2.74 

nc  =  —  ~  =  10  rivets. 
.281 

To  determine  the  number  of  rivets  necessary  for  shear,  the 
required  area  for  shear  is 


A8  —   -     —  =  6.16  square  inches. 
8000 

Each  rivet  is  in  double  shear  and  the  total  area  of  each  rivet  in 
shear  is 

2  A  =  2  X  .442  =  .884  square  inch. 

Therefore,  the  number  of  rivets  required  for  shear  is 

6.16 

n8  =  —  -  =  7  rivets. 
.004 

Since  bearing  requires  10  rivets  that  number  must  be  used. 
The  shearing  stress  then  is  below  the  allowable. 

2.  Calculate  the  unit-stresses  developed  in  a  triple-riveted  lap 
joint  of  a  boiler  4  feet  in  diameter  carrying  no  pounds  per  square 
inch  pressure,  if  the  pitch  is  3  inches,  the  thickness  of  the  plate 
%  inch,  and  the  diameter  of  the  rivets  f  inch.  What  is  the  effi- 
ciency of  the  joint  ? 

The  load  transmitted  through  the  three  rivets  in  the  pitch 
length  of  3  inches  is 

D      QDp        no  X48  X  3 

P  =  -  -  =  7920  pounds. 

The  tension  area  carrying  the  load  is  found  by  assuming  the  diam- 
eter of  the  rivet  hole  the  same  as  that  of  the  rivet. 

•*./<=—  =  —-  =  7040  pounds  per  square  inch. 
At        1.125 


44  RIVETED   JOINTS  [CHAP.  IV 

The  bearing  area  is 

Ac  =  3/d  =  3X^Xf  =  1.125  square  inches. 


.*.     fc  =  -zi-  =  7040  pounds  per  square  inch. 

The  shearing  area  is 

A8  =  3  X  .442  =  1.326  square  inches. 


.*.    fs  =  =5970  pounds  per  square  inch. 

1.326 

By  using  for  the  allowable  unit-stresses  /.  =  8000  pounds  per 
square  inch,  ft  =  12,000  pounds  per  square  inch,  and/c  =  18,000 
pounds  per  square  inch,  the  efficiency  of  the  joint  can  be  calculated. 
The  load  the  unpunched  plate  would  carry  in  the  pitch  length  is 

P  =  12,000  X  I  X  3  =  18,000  pounds. 
The  load  the  punched  plate  will  safely  carry  in  the  pitch  length  is 

Pt  —  12,000  X  1.125  =  I3.5oo  pounds. 
The  load  the  rivets  will  carry  in  compression  is 

Pb  =  18,000  X  1.125  =  20,250  pounds. 
The  load  the  rivets  will  carry  in  shear  is 

Pa  =  8000  X  1.326  =  10,600  pounds. 

The  allowable  load  will  be  the  least  of  these  three  values,  which  is 
10,600  pounds,  and  the  efficiency  is 

10,600 

€  =  ea  =  -      -  =  59  per  cent. 
10,000 

The  efficiencies  for  tension  and  bearing  are  higher.    A  larger 
rivet  would  give  a  higher  efficiency. 


PROBLEMS 

i.  A  column  bracket  consists  of  a  6-inch  X  6-inch  X  f-inch 
angle,  and  is  riveted  to  the  column,  which  is  a  1 2-inch,  30-pound 
channel,  whose  thickness  is  0.513  inch.  It  carries  a  load  of 
20,000  pounds  and  is  riveted  to  the  column  with  5  rivets  I  inch 


PROB.]  RIVETED   JOINTS  45 

in  diameter.     Determine  the  unit-stresses  developed  in  bearing 
and  shear. 

2.  If  two  plates  4  inches  wide  and  f  inch  thick  are  connected 
by  four  f-inch  rivets  in  two  rows,  what  load  will  the  joint  safely 
carry  ? 

3.  Determine  the  required  number  of  rivets  for  a  joint   to 
carry  20,000  pounds,  using  £-inch  plates.    What  is  the  efficiency 
of  the  joint  ? 

4.  Determine  the  required  number  of  rivets  for  a  joint  to  carry 
25,000  pounds,  using  f-inch  plates.     What  is  the  efficiency  of  the 
joint  ? 

5.  Design  an  angle  bracket  to  be  riveted  to  a  column  which 
consists  of  two  1 2-inch,  3o-pound  channels  latticed  together  with 
the   channel   flanges   extending    outwards.     The    bracket    is    to 
support  one  end  of  a  simple  beam  which  carries  a  total  uniform 
load  of  60,000  pounds.     Use  rivets  |  inch  in  diameter.     Neglect 
bending  in  outstanding  leg  of  the  bracket  angle. 

6.  Design  a  splice  to  connect  two  plates  10  inches  wide  and 
f  inch  thick  which  are  subjected  to  a  tension  of  78,000  pounds. 
Use  two  splice  plates  and  rivets  which  are  |  inch  in  diameter. 

7.  Select  two  angles  to  carry  a  tension  load  of  24,500  pounds 
and  determine  the  number  of  rivets  necessary  if  f-inch  gusset 
plates  are  used.     Use  the  allowable  stresses  given  in  example  i. 

Ans.   2  —  2  in.  X  2  in.  X  A  in.  A  •  n  =  5. 

8.  Two  f-inch  plates  are  connected  by  three  f-inch  rivets. 
What  tension  load  will  the  joint  safely  carry  ?     If  two  of  the  rivets 
are  in  one  row  what  should  be  the  width  of  the  plates? 

Ans.    10,600  Ib.  4.1  in. 

9.  In  a  butt  joint  with  a  double  cover  plate  the  main  plates  are 
|  inch  thick  and  the  cover  plates  are  each  ^  inch  thick.     Design 
the  joint  to  take  a  lension  load  of  20  tons.     If  the  load  is  in  com- 
pression how  will  the  design  be  changed  ? 

10.  In  a  boiler  60  inches  in  diameter  carrying  steam  pressure 
at  120  pounds  per  square  inch  the  plate  is  £  inch  thick,  the  rivets 
are  £  inch  in  diameter  and  the  pitch  is  2\  inches.    The  joint 
is  a  double-riveted  lap  joint.    What  are  the  tensile,  shearing,  and 
the  compressive  unit-stresses  coming  in  the  joint?    What  is  the 
efficiency  of  the  joint  ? 


46  RIVETED   JOINTS  [CHAP.  IV 

11.  Determine   the   efficiency   of   a  double-riveted  lap    joint 
where  /  =  ^  inch,  p  =  3^  inches,  and  the  diameter  of  the  rivet 
is  ^|  inch. 

12.  Determine    the    efficiency  of   a   single-riveted,    two-strap 
butt  joint,  if  t  =  f  inch,  p  =  2  inches,  and  the  diameter  of  the 
rivet  is  $  inch. 

13.  Determine  the  pitch  for  a  double- riveted,  two-strap  butt 
joint  in  which  t  =  \  inch,  and  the  diameter  of  the  rivets  is  \\  inch, 
so  that  the  strength  of  the  joint  against  tearing  the  plates  between 
the  rivet  holes  shall  equal  the  compressive  strength  of  the  rivets. 
What  is  the  efficiency  of  this  joint? 

14.  Find  the  thickness  of  plates  for  a  boiler  shell  8  feet  in 
diameter  to  carry  a  pressure  of  160  pounds  per  square  inch,  if 
the  efficiency  of  the  joint  is  80  per  cent,  and  the  stress  in  the  plates 
is  5  tons  per  square  inch. 

15.  Determine  the  efficiency  of  a  single- riveted   lap  joint  if 
t  =  \  inch,  d  =  |  inch,  and  p  =  2  &  inches. 

16.  Calculate  the  efficiency  of   a   double-riveted  lap  joint  if 
/  =  ^  inch,  d  =  i  inch,  and  p  =  3  f  inches. 

17.  Determine  the  pitch  for  a  single-riveted,  two-strap  butt 
joint  in  which  /  =  f  inch  and  d  =  i  ^  inches,  so  that  the  strength 
of  the  joints  against  tearing  the  plates  between  the  rivet  holes 
shall  equal  the  compressive  strength  of  the  rivets.    Determine 
also  the  efficiency  of  the  joint. 

18.  Design  a  double-riveted,  two-strap  butt  joint  for  ^-inch 
plates  and  find  its  efficiency. 


CHAPTER  V 

BEAMS 
EXTERNAL  FLEXURAL  FORCES 

41.  DEFINITIONS.  Flexure  occurs  in  a  member  when 
the  load  has  a  component  normal  to  the  axis  of  the 
member  which  causes  the  member  to  bend.  A  beam  is 
a  bar  subjected  to  flexure.  Usually  the  applied  forces 
are  normal  to  the  axis  of  the  beam,  as  when  a  horizontal 
bar  resting  on  supports  at  its  ends  sustains  vertical  loads 
along  its  length.  However,  the  term  is  also  applicable 
when  the  direction  of  the  applied  forces  is  not  at  right 
angles  to  the  axis.  The  loads  on  a  beam  cause  it  to 
bend  and  thus  produce  internal  stresses  which  resist 
the  bending.  These  stresses  are  called  flexural  stresses. 
The  curve  assumed  by  the  axis  of  the  beam  under  load 
is  the  elastic  curve.  The  following  treatment  considers 
the  beam  to  be  horizontal  and  the  loads  vertical,  but 
with  slight  modifications  it  may  be  adapted  to  beams 
in  any  position  and  with  loads  in  any  direction.  The 
-XT-axis  will  be  taken  to  coincide  with  the  axis  of  the 
beam  before  bending.  The  F-axis  will  be  taken  at  right 
angles  to  the  X-axis  through  the  left  support  or  left  end. 

A  cantilever  beam  is  one  which  has  one  end  free 
and  the  other  end  fixed  in  such  a  manner  that  the  tan- 
gent to  the  elastic  curve  at  the  fixed  end  remains  hori- 
zontal. The  elastic  curve  and  the  beam  itself  may  be 
spoken  of  as  being  horizontal  at  the  fixed  end. 

A  simple  beam  is  one  which  rests  upon  two  end 
supports. 

47 


48         BEAMS  —  EXTERNAL  FLEXURAL   FORCES   [CHAP.  V 

An  overhanging  beam  has  one  or  both  of  its  supports, 
away  from  the  ends  of  the  beam. 

A  continuous  beam  is  one  that  rests,  on  more  than  two- 
supports. 

The  end  of  a  beam  is  said  to  be  fixed  if  it  is  restrained 
in  such  a  manner  that  the  elastic  curve  remains  hori- 
zontal when  the  load  is  applied.  It  follows  from  the 


FIG.  19.    Concentrated  load. 

definitions  above  that  a  cantilever  beam  has  one  sup- 
port, a  simple  beam  has  two  supports,  and  a  continuous 
beam  has  more  than  two  supports. 

42.  METHODS  OF  LOADING  BEAMS.  According  to  the 

distribution  of  the  loads,  a  beam  may  carry  concentrated, 
uniform,  and  nonuniform  or  varying  loads.  When  the 
load  is  transmitted  to  the  beam  through  a  comparatively 
small  area  it  is  said  to  be  concentrated.  Fig.  19  shows 
a  concentrated  load  at  the  center  of  the  span.  If  the 
load  is  distributed  evenly  over  the  beam  it  is  a  uniform 


ART.  43] 


FORCES  ACTING   ON   BEAMS 


load.  Fig,  20  shows  a  load  that  Is  practically  uniform. 
If  the  load  is  distributed  over  the  beam  and  is  not  of 
the  same  intensity  throughout,  it  is  said  to  be  non- 
uniform  or  varying. 

According  to  the  method  of  application,  loads  are 
said  to  be  dead  or  live  loads.  A  dead  load  is  one  that 
the  member  always  supports,  such  as  its  weight  or 


FIG.  20.    Uniform  load. 
(Loaned  by  the  Leonard  Construction  Company,  Chicago.) 

loads  due  to  the  weight  of  other  portions  of  the  struc- 
ture of  which  the  member  is  a  part.  Live  loads  are 
those  that  come  upon  the  member  temporarily,  such 
as  a  train  passing  over  a  bridge,  a  crowd  of  people 
assembled  in  an  auditorium,  or  machinery  or  a  stock  of 
goods.  The  loads  shown  in  Fig.  19  and  20  are  live  loads. 

43.  FORCES  ACTING  ON  A  BEAM  AS  A  WHOLE.    The 

external  forces  acting  on  a  beam  are  in  equilibrium. 
The  loads  supported  by  the  beam  are  usually  known. 


50        BEAMS  —  EXTERNAL   FLEXURAL   FORCES    [CHAP.  V 

The  forces  supporting  the  beam  —  the  reactions  —  may 
not  be  known  at  the  start,  but  when  possible  should 
be  determined  before  making  other  calculations.  The 
loads  and  the  reactions  of  the  supports  form  a  system 
of  parallel  forces.  From  theoretical  mechanics  the  con- 
ditions of  equilibrium  for  such  a  system  are  that  there 
shall  be  no  resultant  force  and  no  resultant  moment. 
These  conditions  are  expressed  in  two  equations: 

2^  =  0,  (i) 

S  M  =  o.  (2) 

These  formulas  are  used  in  determining  the  reactions. 

ILLUSTRATIVE  EXAMPLE 

Let  it  be  required  to  determine  the  reactions  on  a  beam  8  feet 
long,  carrying  a  uniform  load  of  4000  pounds  and  a  concentrated 
load  of  6000  pounds  3  feet  from  the  left  support.  (See  Fig.  21.) 

6000* 
-3- 


4000* 


-8'- 


FlG.    21. 

Let  Ri  be  the  reaction  of  the  left  support  and  R*  the  reaction 
of  the  right  support,  then 

S  F  =  Ri  +  R*  —  4000  —  6000  =  o, 

Ri  +  Rz  =  10,000  pounds, 
2  M a  =  Rz  X  8  -  4000  X  4  -  6000  X  3  =  o, 
R2  =  34000  =  4250  pounds,  and 
Ri  =  10,000  —  4250  =  5750  pounds. 
As  a  check  take  moments  about  the  right  reaction, 
2  Mb  =  6000  X  5  +  4000  X  4  -  Ri  X  8  =  o, 

•Ki  =  5750  pounds. 

The  line  of  action  of  the  distributed  load  is  taken  through  its 
center  of  gravity. 


ART.  44]      EXTERNAL  FORCES  —  INTERNAL  STRESSES      $1 

44.  FORCES  ACTING  ON   A  PORTION  OF  A  BEAM. 

INTERNAL  STRESSES.  In  Fig.  22  is  shown  a  beam  with 
its  loads.  An  effect  of  loading  the  beam  is  to  produce 
internal  stresses  in  the  beam  at  all  sections.  To  study 
the  nature  of  these  stresses  imagine  the  beam  to  be  cut 
along  the  vertical  section  AB.  Then  in  order  that 
equilibrium  in  the  left  portion  of  the  beam  be  main- 
tained with  the  external  loads  and  reactions  acting 
upon  it,  forces  such  as  Vi,  Hi,  and  Hz  must  be  supplied. 


? 

w2 

T  r  T 

Rl 

T 

I'    : 

(a) 

HI 

>H2 

RI 

E 

F 

• 

;    (b) 

IG.    22. 

It  is  evident  that  before  the  section  was  cut  internal 
stresses  must  have  existed  at  this  section,  which  acted 
upon  the  left  portion  of  the  beam.  For  present  purposes 
these  stresses  may  be  considered  to  be  replaced  by 
Vi,  HI,  and  Hz  which  will  later  be  found  to  be  the 
resultants  of  the  internal  stresses.  All  the  forces  shown 
in  Fig.  22  (b)  are  external  with  respect  to  the  left  por- 
tion, but  when  the  whole  beam  is  considered,  FI,  Hi, 
and  H2  are  replaced  by  internal  stresses.  The  reaction 
and  the  loads  on  the  left  portion  of  the  beam  tend  to 


52       BEAMS  —  EXTERNAL  FLEXURAL   FORCES      [CHAP.  V 

cause  motion  upward  (or  downward)  and  rotation  in 
the  clockwise  direction  about  an  axis  in  the  section. 
Both  of  these  tendencies  are  counteracted  by  the  internal 
stresses  at  this  section,  or  by  their  resultants,  FI,  HI, 
and  H2.  The  right  portion  of  the  beam  could  be  treated 
in  a  similar  manner,  but  the  direction  of  FI,  HI,  and 
H2  would  be  opposite  to  that  shown  for  the  left  portion, 
and  the  magnitude  would  be  the  same. 

The  system  of  forces  acting  to  one  side  of  the  sec- 
tion of  the  beam  is  coplanar,  nonparallel,  nonconcurrent. 
The  equations  of  equilibrium  are, 

2  Fy  =  o,  (i) 

2/^  =  0,  (2) 

2  M  =  o.  (3) 

To  satisfy  equation  (i)  the  vertical  resisting  force  FI 
in  Fig.  22  (b)  must  act  downward  (or  upward).  To 
satisfy  equation  (2)  HI  must  equal  H2.  And  to  satisfy 
equation  (3)  the  resisting  forces  must  produce  an  anti- 
clockwise moment,  hence  there  must  be  compression  in 
the  top  fibers  of  the  simple  beam  shown.  In  other 
cases,  as  in  a  cantilever  beam,  tension  may  exist  in  the 
top  and  compression  in  the  bottom  of  the  beam. 

45.  VERTICAL  SHEAR.  From  Art.  44  it  is  seen  that 
the  external  forces  to  the  left  of  the  section  tend  to  cause 
the  left  portion  of  the  beam  to  slip  upward  (or  downward) 
past  the  portion  on  the  right.  Whenever  this  is  the 
case  vertical  shear  is  said  to  exist  at  the  section.  Ver- 
tical shear  is  the  force  that  tends  to  move  the  left  por- 
tion of  a  beam  past  the  right  portion  or  to  cut  the  beam 
along  a  vertical  plane.  The  magnitude  of  this  force  is 
measured  by  the  summation  of  the  vertical  forces  acting 
on  the  beam  to  the  left  of  the  section.  (The  vertical 
forces  acting  on  the  beam  to  the  right  of  the  section 
could  be  used  as  well.) 


ART.  47] 


VERTICAL   SHEAR 


53 


(i) 
-1- 


If   V  represents  the  vertical  shear  at  a  section  the 
distance  x  from  the  left  support, 

in  which  2  RL  represents  the  sum  of  all  the  reactions 
to  the  left  of  the  section,  and  2  WL  is  the  sum  of  all 
the  loads  to  the  left  of  the  section.  The  magnitude 
of  V  is  equal  to  that  of  the  resultant  of  all  the  external 
forces  acting  on  the  left  portion  of  the  beam. 

46.  SIGN  AND  UNIT  OF  VERTICAL  SHEAR.    The  sign 

of  the  vertical  shear  depends  upon  the  relative  values  of 
the  reactions  and  the  loads  to  the  left  of  the  section.  If 
2  RL  is  the  greater,  the 

IA  1 


sign  is  plus,  in  which 
case  the  left  portion 
tends  to  slip  in  the 
upward  or  positive  di- 
rection past  the  right 
portion.  The  unit  of 
vertical  shear  is  the 
same  as  that  used  for 
force,  and  is  usually  the 
pound. 

47.  THE  VALUES  OF 
THE  VERTICAL  SHEAR 
AT  THE  SECTION  AB, 
DISTANT  &  FROM  THE 
LEFT  END  OR  ORIGIN 
FOR  CANTILEVER  AND 
SIMPLE  BEAMS,  (i) 

For  a  cantilever  beam 
with  a  concentrated 
load  W  at  the  end  (Fig. 
23  (i)), 


(2) 


I 

1 

1 

R,=-j 

(3)                       R,-*1 

^«— 

*k 

R<-«{ 

B                 ^                   Ri=M 

FIG.  23. 


54       BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

(2)  For  a  cantilever  beam  with   a   uniform   load   of 
w  pounds  per  inch  of  length  (Fig.  23  (2)), 

V  =  —  wx. 

(3)  For  a  simple  beam  with  a  concentrated  load   W 
in  the  center  of  the  span  (Fig.  23  (3)), 

W 

V  =  —  to  the  left  of  the  center 
2 

or  V  = to  the  right  of  the  center. 

(4)  For  a  simple  beam  with  a  uniform  load  of  w  pounds 
per  inch  of  length  (Fig.  23  (4)), 

T7  ™l 

V  = wx. 

2 

ILLUSTRATIVE  EXAMPLES 

1.  If  a  cantilever  beam  of  8-ft.  span  carries  a  load  of  500  pounds 
at  the  free  end,  the  value  of  the  vertical  shear  at  any  section  is 
—  500  pounds.     The  left  reaction  is  zero  and  the  load  to  the  left 
is  500  pounds;  therefore  V  =  o  —  500  =  —  500  pounds. 

2.  For  a  simple  beam  of  lo-ft.  span  carrying  a  uniform  load 
of  6000  pounds,  the  left  reaction  is  3000  pounds.      If  x  is  ex- 
pressed in  inches  w  =  6000  -f-  120  =  50  pounds  per  inch  and  the 
expression  for  the  vertical  shear  is 

V  =  3000  —  50  x. 

By  substituting  values  of  x  in  this  equation  we  obtain  values  of 
the  vertical  shear  for  the  sections  considered;  thus  at  25  inches, 
V  =  3000  —  1250  =  1750  pounds.  At  the  distance  6  feet  or 
72  inches  from  the  left  support  the  vertical  shear  is 

V  ="3000  —  3600  =  —  600  pounds. 

48.  LOAD  AND  SHEAR  DIAGRAMS.    It  is  convenient 

and  useful  to  indicate  the  value  of  the  load  and  the  shear 
at  all  points  along  the  beam  by  means  of  vertical  dis- 
tances measured  from  a  horizontal  axis.  These  verti- 


ART.  48]  LOAD   AND   SHEAR  DIAGRAMS  55 

cal  distances  are  called  ordinates.  With  the  length  of 
the  beam  and  the  loading  known,  the  axis  OX,  Fig.  24, 
can  be  drawn  to  scale  to  represent  the  length  of  the 
beam.  Thus,  if  the  beam  is  12  feet  or  144  inches  long, 
and,  if  it  is  convenient  to  make  the  axis  OX  3  inches 
long,  the  scale  of  the  length  would  be  I  inch  equals 
48  inches.  From  the  axis  in  Fig.  24  at  any  point  A  a 


FIG.  24. 

length  AB  can  be  erected  perpendicular  to  OX  to  repre- 
sent the  intensity  of  load  at  that  point.  If  the  load 
at  that  point  is  240  pounds  per  foot  of  length  or  20  pounds 
per  inch  of  length,  the  ordinate  AB  can  be  made  I  inch, 
in  which  case  I  inch  would  represent  a  load  of  100  pounds 
per  inch.  In  the  same  way  the  value  of  the  intensity 
of  load  at  all  other  points  along  the  beam  can  be  repre- 
sented by  ordinates.  The  continuous  curve  connecting 
the  ends  of  these  ordinates  is  called  the  load  curve, 
and  the  whole  figure  the  load  diagram,  because  the  inten- 
sities of  the  load  are  shown  at  every  point  along  the 
beam.  The  locations  of  the  concentrated  loads  are 
indicated  by  arrows.  Positive  values  are  measured  to 
the  right  of  and  up  from  the  origin  0,  negative  to  the 
left  of  and  down  from  the  origin  O.  Loads  act  down 
and  consequently  are  negative. 

Knowing  the  loading,  the  reactions  can  be  calculated 
and  the  values  of  the  vertical  shear  can  be  obtained 
for  all  sections  along  the  beam  by  the  formula 
V  = 


56       BEAMS  —  EXTERNAL  FLEXURAL   FORCES      [CHAP.  V 


By  a  procedure  similar  to  that  used  in  making  the  load 
diagram  the  values  of  the  vertical  shear  at  all  points 
along  the  length  of  the  beam  can  be  indicated  by  ordi- 
nates  from  a  reference  axis  OX.  Thus,  Fig.  25  is  the 

shear  diagram  for  the 
loading  shown  in  Fig. 
24.  In  the  shear  dia- 
gram the  distance  from 
the  axis  OX  to  the 
curve  at  any  point  A' 
shows  the  value  A'B' 
of  the  shearing  force 
for  that  section  of  the 


FIG.  25. 


beam. 


(a) 


49.  RELATION  BETWEEN  THE  LOAD  AND  THE  SHEAR. 

For  distributed  loads  there  is  a  definite  relation  between 

the  load  and  the  shear. 

To  deduce  this  relation 

let   Fig.   26    (a)    be    the 

load  diagram  and  Fig.  26 

(b)    the    shear    diagram 

for  a  given  case.     Let  u 

be  a  small  length  (called 

an    element    of    length) 

measured  along  the  OX 

axis,  and  let  w  be  the 

average  load  per  unit  of 

length  over  this  portion. 

Let   x   be    the   distance 

from  the  left  support  or 

origin,  and  V  the  verti-  FlG-  26- 

cal  shear  at  the  left  end  of  the  element  of  length.     The 

load  over  this  length  is  wu,  and  is  equal  to  the  small 

shaded  area  in  the  load  diagram.     Then  the  difference 


ART.  51]       THE   LOAD   AND   SHEAR   DIAGRAMS 


57 


in  the  vertical  shear  at  the  two  sides  of  the  small  length 
is 

v  =  wu. 

50.  THE  RATE  OF  CHANGE  OF  THE  VERTICAL  SHEAR. 

The  rate  at  which  the  vertical  shear  changes  is  equal  to 
the  amount  of  change  divided  by  the  length  in  which 
the  change  is  made,  and  is 


wu 


=  w. 


u  may  be  taken  so  small  that  A  C  in  Fig.  26  (b)  approaches 
a  straight  line.  When  u  is  made  indefinitely  small  AC 
coincides  with  the  tangent  to  the  shear  curve.  It  is 

also  seen  that  -  equals  tan  ACB,  which  represents  the 

slope  of  the  shear  curve  at  the  given  section  when  u  is 
indefinitely  small,  w  is  the  intensity  of  the  load  at 
that  section.  Therefore,  the  rate  of  change  of  the 
vertical  shear  at  any  sec- 
tion of  a  beam  equals  the  R 
intensity  of  the  load  at 
that  section.  It  is  also 
represented  by  the  slope 
of  the  shear  curve  at 
that  section. 


51.  RELATION  BE- 
TWEEN THE  LOAD  AND 
SHEAR  DIAGRAMS.  The 

relation  given  in  Art.  49 
affords  a  convenient 
graphical  method  of  de- 
termining the  change  in 
the  vertical  shear  between  any  two  sections  AB  and  CD 
of  a  beam,  Fig.  27.  Divide  the  length  AC  into  several 


FIG.  27. 


58       BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

small  lengths.  Then  v\  equals  area  ABn,  vz  equals  area 
1 122,  va  equals  area  2233,  etc.  The  total  change  vr  be- 
tween the  two  sections  is  the  summation  of  all  the  w's 
between  them.  This  summation  is  equal  to  the  area 
under  the  load  curve  between  the  sections,  which  is  the 
total  load  between  the  two  sections.  Therefore,  the 
change  in  the  vertical  shear  between  any  two  sections 
of  a  beam  equals  the  area  under  the  load  curve  between 
the  two  sections.  (Distributed  loads.) 

52.  BENDING  MOMENT.    Moment  of  a  force  is  the 
product   of   the   force   and   the   perpendicular   distance 

from  the  line  of  action  of 
the  force  to  the  origin  of 
moments.  The  moment 
measures  the  tendency  of 
the  force  to  cause  the  ob- 
ject acted  upon  by  the 

force  to  turn  about  an  axis 
FIG.  28.  ,,  1^1  •  •  r 

through  the  origin  of  mo- 
ments. Fig.  28  indicates  a  force  F  acting  upon  an  ob- 
ject which  it  tends  to  turn  about  the  point  0.  The 
moment  in  this  case  is  Fa  where  a  is  the  perpendicular 
distance  from  the  origin  0  to  the  line  of  action  of  the  force. 
For  a  section  of  a  beam  the  bending  moment  is  the  sum 
of  the  moments  of  all  external  forces  acting  on  the  beam 
to  one  side  of  the  section  about  an  axis  in  the  section.  A 
result  of  a  bending  moment  is  a  tendency  to  cause  the 
portion  of  the  beam  considered  to  rotate  about  an  axis 
in  the  section.  In  determining  the  bending  moment, 
the  portion  of  the  beam  to  the  right  or  the  one  to  the 
left  of  the  section  may  be  used  with  the  same  results. 
It  is  the  common  convention  to  use  the  left  portion  of 
the  beam  and  this  convention  will  be  followed  in  this 
book.  The  bending  moment  at  any  section  then  is  equal 


ART.  53]    SIGN  AND   UNIT  OF  BENDING  MOMENT         59 

to  the  summation  of  the  moments  of  all  the  forces  acting 
upon  the  beam  to  the  left  of  the  section  about  an  axis 
in  the  section.  Since 

the  external  forces  are  T1       Y2      ^        Y3      Y4 

made  up  of  the  reac- 
tions and   loads,    the 
bending    moment    M    - 
equals    the    moments  Ri 
of  the  reactions  minus 
the   moments   of   the 
loads  to  the  left  of  the  section,  or  expressed  in  a  formula 
(see  Fig.  29), 

M  =  2RLX-  2W(x-p), 

in  which  %RLX  is  the  summation  of  the  moments  of 
the  reactions,  and  2W (x  —  p)  is  the  summation  of  the 
moments  of  the  loads  to  the  left  of  the  section. 

53.  SIGN  AND  UNIT  OF  BENDING  MOMENT.  When 
there  is  compression  in  the  top  fibers  of  the  beam  at  a 
section  the  bending  moment  is  positive  at  the  section; 
when  there  is  tension  in  the  top  fibers  the  sign  of  the 
bending  moment  is  negative.  The  radius  of  curvature 
of  the  elastic  curve  is  positive  for  a  positive  bending 
moment  and  negative  for  a  negative  bending  moment. 
The  sign  'of  the  bending  moment  due  to  a  force  is  posi- 
tive when  the  force  itself  would  produce  compression 
in  the  top  fibers.  Hence  in  all  cases  the  bending 
moment  of  a  reaction  is  positive,  and  that  of  a  load  is 
negative.  The  unit  in  which  the  bending  moment 
is  measured  will  depend  upon  the  units  of  force  and 
length  employed.  The  pound-inch*  is  in  most  common 
use  for  beams  and  will  be  employed  here. 

*  The  term  "  inch-pound,"  which  is  also  used  for  the  unit  of  moment, 
does  not  make  a  distinction  between  the  unit  of  work  and  the  unit  of 
moment. 


60      BEAMS  —  EXTERNAL   FLEXURAL  FORCES      [CHAP.  V 


Ar,_W 

Rrir 


IB 


(i) 
-I- 


54.  THE  VALUES  OF  THE  BENDING  MOMENT  AT 
THE  SECTION  AB,  DISTANT  a?  FROM  THE  LEFT  END 
OR  ORIGIN  FOR  CANTILEVER  AND  SIMPLE  BEAMS. 

(1)  Cantilever   beam, 
concentrated  load  W  at 
the  end 

M  =  -Wx. 

Left  reaction  is  zero 
(Fig.  30  (i)). 

(2)  Cantilever   beam, 
uniformly   distributed 
load    of    w  pounds  per 
inch 

ir  X  ^ 

M  =  —  wx-  =  —  w—  • 

2  2 

The  left  reaction  is  zero. 
The  load  to  the  left  of 
the  section  is  wx  and  its 


(2) 


W 


arm  is  -  (Fig.  30  (2)). 

(3)  Simple  beam,   con- 
centrated load  Wat  center. 

W 

The  left  reaction  is  —  » 
2 


—  x  for  the  left  half, 

2 


(3) 


FIG.  30. 


M  = 
and 


=  Wl    Wx 

2  2 

for   the    right   half    (Fig. 
30(3)). 


ART.  55]  BENDING-MOMENT  DIAGRAMS 


61 


(4)  Simple  beam,  uni- 
form load  of  w  pounds 
per  inch. 

The  left  reaction  is  — 

2 

,,      wl  x 

M  =  —  x  —  wx  - 
2  2 

_  wlx      wx2 

2  2 

The  load  to  the  left  of  the 
section  is  wx  and  its  arm 

is      (Fig.  30  (4)). 


M 


FIG.  30. 


ILLUSTRATIVE  EXAMPLES 

1.  For  a  cantilever  beam  of  g-it.  span  carrying  a  load  of  2000 
pounds  at  the  free  end  the  bending  moment  at  the  section  distant 
x  from  the  free  end  is  M  =  —  2000  x.     If  x  is  50  inches,  M  = 
—  2000  X  50  =  —  100,000  Ib.-in. 

2.  For  a  simple  beam  of  2o-ft.  span  with  a  concentrated  load 
of  10,000  pounds  at  the  center  the  left  reaction  is  5000  pounds 
and  the  bending  moment  at  8  feet  or  96  inches  is 

M ge  =  5000  X  96  =  480,000  Ib.-in. 
At]  1 6  feet  the  bending  moment  is 

Mm  =  5°°o  X  192  —  10,000  X  72  =  240,000  Ib.-in. 

3.  For  a  simple  beam  of  i2-ft.  span  carrying  a  uniform  load  of 
100  pounds  per  inch  Rt  =  7200  pounds  and  the  bending  moment 
at  the  center  is 

IPO  X  72  X  72 
M 72  =  7200  X  72  -   -  -  =  259,200  Ib.-in. 


55.  BENDING-MOMENT   DIAGRAMS.    The  bending- 

moment  diagram  shows  the  value  of  the  bending  mo- 
ment at  all  points  along  the  beam.  Fig.  31  (d)  is 
such  a  diagram.  OX  represents  to  scale  the  length  of 
the  beam,  and  the  ordinate  M  represents  the  bending 


62      BEAMS  —  EXTERNAL   FLEXURAL  FORCES      [CHAP.  V 


moment  at  the  section  AB.     The  values  of  the  bending 
moment  may  be  calculated  by  the  formula 


A      C 


B    6 


(d) 

FIG.  31. 


or  by  the  method  of  the 
following  article. 

56.  RELATION  BE- 
TWEEN THE  VERTICAL 
SHEAR  AND  THE  BEND- 
ING MOMENT.  At  a 

section  distant  x  from 
the  left  support  the 
bending  moment  is  M, 
and  at  a  section  distant 
x  +  u  it  is  M  +  m,  where 
u  is  an  element  of  length 
along  the  OX-  axis  and 
m  is  the  difference  in  the 
bending  moment  at  the 
two  sections  .  I  n  the  free- 
body  diagram  of  the 
element  of  length  of  the 
beam  between  the  two 
sections  shown  in  Fig. 
31  (a)  M  is  the  bending 
moment  of  the  external 
forces  to  the  left  of  the 
section  AB.  This  mo- 
ment is  transferred  to 


the  element  of  the  beam,  and  the  bending  moment  trans- 
ferred from  this  element  to  the  right  portion  is  M  +  m. 
The  increase  m  is  due  to  the  external  forces  acting  on  the 
small  portion  of  the  beam.  The  external  forces  with 
respect  to  this  portion  acting  upon  it  are  V  the  vertical 
shear  at  the  left,  wu  the  weight,  where  w  is  the  weight 


ART.  57]      CHANGE  OF  THE   BENDING   MOMENT  63 

per  unit  of  length,  and  V  +  v  the  vertical  shear  at  the 
right,  as  shown  in  the  diagram.  Taking  moments  about 
an  axis  in  the  section  CD,  there  results 

u  uz 

M  +  m  =  M  -\-Vu  —  wu-  =  M  +  Vu  —  w  — ; 

• 

m  =  Vu  —  w  —  • 

2 

u  may  be  taken  so  small  that  the  load  over  this  elemen- 
tary length  may  be  considered  uniform.  From  Fig.  31 
(c)  it  is  seen  that  Vu  is  the  area  of  the  rectangle  EFHI 


u2 
and  w  —  is  that  of  the  triangle  EGI. 

Therefore,  m  =  EFHI  —  EGI,  which  is  equal  to  the  area 
between  the  axis  and  the  shear  curve.  If  the  sections 
are  taken  far  apart  the  distance  between  the  two  sec- 
tions should  be  divided  into  a  large  number  of  elementary 
lengths,  and  the  total  change  in  the  bending  moment 
will  equal  the  summation  of  all  the  elementary  changes. 
This  leads  to  the  conclusion  that  the  change  in  the  bend- 
ing moment  between  two  sections  equals  the  area  under 
the  shear  curve  between  the  two  sections.  Having 
the  vertical  shear  diagram  drawn  and  knowing  the  value 
of  the  bending  moment  for  any  section  of  the  beam, 
that  for  any  other  section  may  be  obtained  by  getting 
the  area  under  the  shear  curve  between  the  two  sections 
and  adding  it  to  the  known  moment.  Areas  above  the 
axis  are  positive  and  those  below  are  negative. 

57.  THE  RATE  OF  CHANGE  OF  THE  BENDING  MOMENT. 

From  the  equation  for  the  change  of  the  bending  moment 

u? 
between   two    sections,   m  =  Vu  —  w  —  the   expression 

for  the  rate  at  which  the  bending  moment  changes  at 
any  section  may  be  deduced  by  allowing  u  to  become 


64  BEAMS  —  EXTERNAL  FLEXURAL  FORCES  [CHAP.  V 
so  small  that  the  two  sections  AB  and  CD  will  be  con- 
secutive ;  then  w  —  is  so  small  compared  to  Vu  that  it 

may  be  considered  equal  to  zero.  Therefore  the  rate 
of  change  is 

m  _  Vu  _ 

w        u 

That  is,  the  rate  of  change  of  the  bending  moment  at 
any  section  equals  the  vertical  shear  at  that  section. 

58.  THE  MAXIMUM  VERTICAL  SHEAR  AND  BENDING 

MOMENT.  The  greatest  shearing  stress  will  be  at  the 
section  for  which  the  vertical  shear  is  the  greatest  and 
the  greatest  tensile  and  compressive  moment  stresses  will 
be  where  the  bending  moment  is  the  greatest.  In  any 
kind  of  beam  the  greatest  shear  occurs  just  to  one  side 
of  a  support.  Because  beams  usually  fail  at  the  sec- 
tion of  maximum  bending  moment,  that  section  is 
called  the  danger  section.  From  the  relation  existing 

between  the  shear  area  and 
the  bending  moment  there 
is  found  a  simple  method 
of  locating  the  danger  sec- 
tion. By  Art.  56  the  area 
under  the  shear  curve  be- 
tween two  sections  repre- 
sents the  change  in  the 
bending  moment  between 
those  two  sections.  As  we 
go  along  the  beam  the 
(d)  bending  moment  increases 

FlG  as  long  as  the  shear  area 

is  positive,  and  when  the 

area  becomes  negative  the  moment  grows  less  (Fig.  32). 
Shear  area  above  the  axis  is  positive  and  that  below  is 


ART.  59] 


LOAD,   SHEAR,   AND   MOMENT 


negative,  the  sign  of  the  shear  changing  where  the  shear 
curve  crosses  the  axis.  Therefore,  the  maximum  bending 
moment  in  a  beam  occurs  where  the  shear  curve  crosses 
the  axis,  i.e.,  where  the  vertical  shear  is  zero.  This 
point  may  be  obtained  by  plotting  the  shear  curve.  It 
may  also  be  obtained  from  the  equation  of  the  vertical 
shear  for  the  portion  of  the  beam  in  which  the  shear 
passes  through  zero,  by  equating]  V  to  zero  and  solving 
for  x.  Note  that  the  equation  representing  V  must  be 
for  the  portion  of  the  beam  in  which  the  shear  actually 
does  pass  through  zero.  Whenever  the  shear  passes 
through  zero  at  a  concentrated  load,  the  maximum  bend- 
ing moment  will  be  under  that  load. 

Since  the  bending  moment  is  zero  at  both  supports  of  a 
simple  beam  the  shear  area  above  the  axis  is  equal  to  that 
below  the  axis. 

59.  LOAD,  SHEAR,  AND  MOMENT  DIAGRAMS  FOR 
CANTILEVER  AND  SIMPLE  BEAMS.  MAXIMUM  SHEAR 
AND  MOMENT.  For  all 

cases  let  x  be  the  dis- 
tance from  the  origin 
to  the  section  AB  con- 
sidered, W  the  total 
load  on  the  beam,  /  the 
span,  Vm  the  maximum  0 
shear,  and  Mm  the  max- 
imum moment. 

(i)  For  a  cantilever 
beam  with   a   load   at  ol 
the  end  (Fig.  33), 

M  =  -Wx. 

The  area  in  shear  dia- 
gram  is  negative  and 


66      BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

equals  —  Wx.  The  maximum  moment  occurs  at  the 
wall  and  equals  the  entire  area  in  the  shear  diagram 
which  is 

Mm  =  -  Wl. 

(2)  For    a    cantilever    beam    with    a    uniform    load 
(Fig.  34) .     The  load  per  unit  of  length  is 

W 


F  =  -y*,     Vm=-W, 

which  comes  at  the  wall. 
t 


_         r2    __Wtf 

2s    '  T  2" 

This  is  equal  to  the  shear  area  to  the  left  of  the  section. 


ART.  59]  '         LOAD,   SHEAR,  AND  MOMENT 


67 


The  maximum  moment  comes  at  the  wall  and  equals 
the  entire  area  under  the  shear  curve  and  is 


2  2 

(3)  For  a  simple  beam  with  a  concentrated  load  at  the 
center  of  the  span  (Fig.  35)  , 

w 


V=—to  the  left  of  the  load, 

W 
V  = to  the  right  of  the  load, 


68      BEAMS  —  EXTERNAL  FLEXURAL  FORCES      [CHAP.  V 


Vm   = 


w 


w 


M  =  —  x  to  the  left  of  the  load 


W 
— 


w/         l\ 
x-  W(x  -  -j  = 


^ 
-  x) 


to  the  right  of  the  load.     The  shear  passes  through  zero 
under  the  load;  therefore  the  maximum  moment  occurs 

at  the  center  of  the  beam. 

The  area  under  the  shear 

curve  to  the  left  of  the 

center  is 

W     l  =  Wl 

22          4   ' 

,,        Wl 

Mm=—' 

4 

(4)  For  a  simple  beam 
with  a  uniform  load  (Fig. 
x  36),  the  load  per  unit  of 

W 
length  is  —  w  =  — •=-• 

=  wl  =  W 

2  2 

wl                W     W 
V  = wx  = j-Xj 

2  2  I 

v  •-—  • 

m         2 

The   average   ordinate   in 
the  shear  area  to  the  sec- 
tion at  the  distance  x  from  the  left  end  is 


wl 


wl 

_ 


wx 

_ 


ART.  59]  LOAD,   SHEAR,  AND   MOMENT 


69 


therefore  the  area  under  the  curve  to  the  section  equals 

wlx       wx2 


M 


2  2 

wlx       wx2       Wx 


2  2  2  2  / 

The  shear  passes  through  zero  at  the  center  of  the  span ; 
therefore  the  maximum  moment  occurs  at  that  section 
and  equals  the  shear  area  to  the  left  of  the  center. 


. 

m~222~8  8 

(5)  For  a  simple  beam  with  a  concentrated  load  at 


JM 

FIG.  37. 

any  point  distant  kl  from  the  left  support,  in  which  k 
is  less  than  i  (Fig.  37), 

Ri=  W(i  -k), 

V  =  W(i  -  k)  to  the  left  of  load, 
V  =  -  Wk  to  the  right  of  load, 
M  =  W(i  -  k)  x  to  the  left  of  load, 
M=  W(i  -  k)x  -W(x-kl) 

=  Wk(l-  x)  to  the  right  of  load. 


70      BEAMS  —  EXTERNAL  FLEXURAL   FORCES      [CHAP.  V 


The  shear  passes  through  zero  under  the  load.    The  area 
to  the  load  is 


(6)  A  uniform  and  a  concentrated  load  on  a  simple 
beam  (Fig.  38).     For  cases  of  this  kind  the  shear  will  be 

a  maximum  at  one  of 
the  supports  and  will 
pass  through  zero  under 
the  concentrated  load  or 
between  that  load  and 
the  center  of  the  beam. 
The  point  at  which  it 
passes  through  zero  de- 
x  pends  on  the  ratio  of 
the  loads  and  the  posi- 
tion of  the  concentrated 
load.  The  shear  curve 
should  be  plotted  to 
obtain  that  point,  then 
the  area  under  that 
curve  above  the  axis 
calculated  for  the  maxi- 
mum bending  moment. 

(7)  For   several    concentrated    loads    the    maximum 
shear  occurs  at  one  of  the  reactions  and  it  passes  through 
zero  at  one  of  the  loads   (Fig.  39).     The  point  of  zero 
shear  may  be  obtained   by  plotting  the  shear  curve. 
Then  the  maximum  moment  may  be  obtained  by  cal- 
culating the  area  in  the  shear  diagram  above  the  axis. 

(8)  For  the  case  of  several  concentrated  loads  and  a 
uniform  load  the  shear  may  pass  through  zero  at  one 
of  the  concentrated  loads  or  between  any  two  of  them, 
the  position  of  zero   shear   depending   upon   the  rela- 
tive values  of  the  loads  and  their  positions  (Fig.  40). 


ART.  59]  LOAD,   SHEAR,   AND   MOMENT 


LJ  1 


(a) 


(b) 


(c) 

FIG.  39. 

r r 


W  vf/in. 


(a) 


(6) 


(c) 


FIG.  40. 


72      BEAMS  —  EXTERNAL  FLEXURAL  FORCES      [CHAP.  V 

The  shear  diagram  should  be  plotted  to  locate  the  section 
of  zero  shear.  To  find  the  exact  location  when  the  sec- 
tion of  zero  shear  falls  between  two  loads  the  shear 
formula  for  a  section  between  these  two  loads  may  be 
equated  _to  zero  and  the  corresponding  value  of  x  deter- 
mined. Then  the  moment  equation  applied  for  that 
point  will  give  the  value  of  the  maximum  moment,  or 
the  area  in  the  shear  diagram  above  the  axis  may  be 
obtained  for  the  maximum  moment. 

(9)  For  the  case  of  a  beam  overhanging  one  or  both 
supports   the   general   principles   hold    (Fig.   41).     The 


w, 


(a) 


(6) 


|W3 


IR, 


maximum  shear  generally  is  not  equal  to  a  reaction,  but 
it  may  be  obtained  by  the  application  of  the  general 
equation  for  shear.  It  will  come  just  to  one  side  of  a 
support.  The  shear  passes  through  zero  at  the  supports, 
and  also  between  them,  consequently  the  maximum 


ART.  60]        CANTILEVER   AND    SIMPLE   BEAMS 


73 


bending  moment  may  occur  at  any  one  of  the  three 
points.  The  moment  must  be  obtained  for  all  three  and 
the  greatest  one  taken  for  the  maximum. 

60.  RELATIVE  STRENGTH  OF  CANTILEVER  AND  SIMPLE 

BEAMS.  The  shearing  unit-stresses  developed  in  a 
beam  are  directly  proportional  to  the  vertical  shear, 
and  the  unit-stresses  resisting  the  bending  moment  are 
directly  proportional  to  the  bending  moment;  therefore 
the  shearing  strength  of  a  certain  type  of  beam  is  in- 
versely proportional  to  the  maximum  vertical  shear 
developed  in  that  type,  and  the  strength  to  resist  bend- 
ing is  inversely  proportional  to  the  maximum  bending 
moment  developed.  Table  12  gives  the  values  of  the 
maximum  vertical  shear  and  the  maximum  bending  mo- 
ment, and  the  relative  strengths  in  shear  and  bending 
for  simple  and  cantilever  beams. 

TABLE  12 

RELATIVE   STRENGTHS  IN   SHEAR  AND   BENDING 


Kind  of  beam. 

Maximum 
vertical 
shear. 

Maximum 
bending 
moment 

Relative 
strength  in 
shear. 

Relative 
strength  in 
bending. 

1                            I 

TT/ 

TT// 

<c                            I                        $' 

W 

I 

I 

m 

Wl 

w/7  y  in.                 fy 

w 

2 

I 

2 

f 

—  y*-f 

i 

W 

WL 

\          •  •      f 

2 

4 

4 

w/zyin. 

w 

Wl 

Q 

2 

8 

\                         •  f 

74     BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 


ILLUSTRATIVE  EXAMPLES 

Calculate  the  maximum  bending  moment  developed  in  a  beam 
of  1 2-f  t.  span  carrying  a  load  of  800  pounds,  (a)  when  used  as  a 
cantilever  with  the  load  concentrated  at  the  end;  (b)  if  the  load 
is  uniformly  distributed  on  the  cantilever;  (c)  when  used  as  a 
simple  beam  with  the  load  concentrated  at  the  center;  (d)  if  the 
load  is  uniformly  distributed  over  the  simple  beam. 

(a)  Mm  =  —800  X  12  X  12  =  —115,200  Ib.-in.  at  the  wall. 

(b)  Mm  =  8o°  x  I2  x  I2  =  - 57,600  Ib.-in.  at  the  wall. 

(c)  Mm  =  8°°  X  I2  X  *-  =  28,800  Ib.-in.  at  the  center. 

4 

(d)  Mm  =  8°°   XI2   XI2    = 


61.  MOVING  CONCENTRATED  LOADS  ON  A  BEAM. 

When  several  concentrated  loads  pass  over  a  beam  the 
beam  must  be  strong  enough  to  take  the  greatest  shear 
and  the  greatest  bending  moment  caused  by  the  loads. 
Hence  it  is  necessary  to  determine  the  maximum  shear 
and  moment  developed  by  the  system  of  loads.  The 
greatest  vertical  shear  will  be  developed  at  the-  support 
when  one  of  the  loads  is  very  near  it. 

The  greatest  bending  moment  will  occur  under  one 
of  the  loads,  since  the  moment  curve  for  any  position 
of  the  loads  consists  of  a  series  of  straight  lines.  For 
simple  beams,  when  one  of  the  loads  is  over  a  support 
the  bending  moment  under  that  load  will  be  zero.  As 
the  system  of  loads  passes  over  the  beam  the  bending 
moment  under  each  load  will  increase  from  zero,  when 
at  the  support,  to  a  maximum  value  when  it  is  at  some 
point  between  the  supports,  and  then  decrease  to  zero 
at  the  other  support.  Let  us  determine  the  position 
of  the  system  of  loads  that  will  give  the  maximum  bend- 
ing moment  under  a  particular  load,  as,  for  example, 


ART.  61]  MOVING   LOADS   ON  BEAMS  75 

Wz  in  Fig.  42.     Let  R  be  the  resultant  of  all  the  loads 
and  its  line  of  action  be  at  the  distance  x  from  load  W% 


w2  w3!  w4    w5 


FIG.  42. 

and  xi  from  the  right  support.     Let  the  load  be  at  the 
distance  x  from  the  left  support;  then 

Rxi  — 

RI  =  —j-  and  xi  =  I  —  x  —  x. 

The  moment  under  the  load  then  is 


i  o  — 

=  -rXl  --  TX    --  TXX  — 
III 


By  algebra  it  can  be  shown*  that  the  value  of  x  to 
give  the  greatest  value  of  a  function  of  the  form 

ax2  +  bx  -\-  c  is 

By  substituting  in  this  formula  for  x, 

Rn 

-  (I  -  x)       .      _ 


x  = 


2T 


2  x  =  I  —  x  or  transposing  x, 

x  =  I  —  x  —  x  =  Xi. 

This  shows  that  the  position  of  one  of  a  system  of 
moving  concentrated  loads,  when  the  greatest  bending 
moment  occurs  under  that  load,  is  such  that  its  distance 
from  the  left  support  is  tl^e  same  as  the  distance  of  the 
center  of  gravity  of  all  the  loads  from  the  right  support. 
*  See  "Higher  Algebra,"  by  John  F.  Downey,  page  245. 


76      BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

In  order  to  obtain  the  maximum  bending  moment 
produced  by  the  system  of  moving  loads  the  maximum 
should  be  determined  for  each  load  and  the  highest 
value  taken. 

ILLUSTRATIVE  EXAMPLE 

Let  it  be  required  to  obtain  the  maximum  bending  moment 
produced  in  a  beam  of  21 -ft.  span  by  a  system  of  three  moving 
loads  of  4  tons,  2  tons,  and  3  tons  spaced  3  feet  and  4  feet  apart 
respectively.  (Fig.  43.) 

The  resultant  is  9  tons.  Its  line  of  action  with  respect  to  the 
loads  may  be  obtained  by  taking  moments  about  any  load.  The 

4T  2T  3T 


-21- 


i 


FIG.  43. 


line  of  action  of  the  resultant  passes  through  the  2- ton  load. 
For  the  maximum  moment  coming  under  the  4- ton  load  its 
distance  from  the  left  end  must  be  the  same  as  the  distance  of 
the  resultant  of  the  three  loads  from  the  right  end.  x  for  this 
load  is  3  feet;  therefore,  the  distance  of  the  load  from  the  left 

support  is  x  =  —  — -  =  9  feet.     Placing  the  loads  in  this 

position,  obtaining  the  left  reaction  by  taking  moments  about  the 
right  support,  and  using  the  resultant  instead  of  each  load  sepa- 
rately, there  results 

Ri  =  —  X  9  X  2000  =  7710  pounds. 

The  moment  under  this  load  is 

Mi  =  7710X9X12  =  833,000  pound-inches. 

The  maximum  moment  will  come  under  the  2-ton  load  when 
it  is  at  the  center  of  the  beam,  since  x  for  this  load  is  zero;  then 

R!  =4.5  X  2000  =  9000  pounds. 
Mz=  (9000 X  io|  -  8000  X  3)  12  =  846,000  Ib.-in. 


EXAM.]     BEAMS  —  EXTERNAL  FLEXURAL  FORCES         77 
For  the  3-ton  load  x  =  —4  feet, 


T  2    £ 

.Ri  =  —  —  X  9  X  2000  =  10,710  pounds. 
21.0 

M3  =  (10,710  X  12.5  —  8000  X  7  —  4000  X  4)12  =  744,000  Ib.-in. 

The  maximum  moment  for  this  system  of  loads  then  is  devel- 
oped under  the  2-ton  load  and  is  846,000  pound-inches. 


EXAMPLES 

i.  Given  a  simple  beam,  i5-ft.  span,  with  a  load  varying 
uniformly  from  zero  at  the  left  end  to  1000  pounds  per  lineal 
foot  at  the  right  end,  and  a  concentrated  load  of  9000  pounds 
6  feet  from  the  left  end.  Determine  the  reactions.  (See  Fig,  44.) 


FIG.  44- 

The  average  distributed  load  is  (o  +  1000)  -r-  2  =  500  pounds 
per  foot.     The  total  distributed  load  is  500  X  15  =  7500  pounds. 
The  center  of  gravity  of  this  load  is  f  X  15  =  10  feet  from  the 
left  support.     Taking  moments  about  the  right  support, 
2  MB  =  9000  X  9  + 7500  X  5  -  R!  X  15  =  o, 

„        9000 X  9  +  7500  X  ^ 
.'.     Ri  =  -  -  =  7900  pounds. 

Taking  the  summation  of  the  forces  to  obtain  R2, 

2  F  =  7900  +  R2  -  9000-  7500  =  o;      /.     R2  =  8600  pounds. 


78      BEAMS  —  EXTERNAL  FLEXURAL  FORCES      [CHAP.  V 

For  checking  on  R-2  take  moments  about  Ri, 

2  MA  =  RzX  15  —  9000 x  6  -  7500 x  10  =  o; 

D      0000X6  —  7500X10      0, 
/.    R2  =  -  -  =  8600  pounds. 

2.  Calculate  the  shear  and  bending  moment  at  sections  2  feet 
apart  and  draw  the  shear  and  moment  diagrams  for  a  simple 
beam  of  i4-ft.  span  with  a  uniform  load  of  14,000  pounds  and 
a  concentrated  load  of  7000  pounds  5  feet  from  left  support. 

RI  =  r7?  X  14,000  +  T\  X  7000  =  7000  +  45oo  =  11,500  pounds. 

R2  —  21,000  —  11,500  =  9500  pounds. 
Check  R2  =  T?  X  14,000  +  t\  X  7000  =  9500  pounds. 

From  the  definition         V  =  R!  -  2  WL, 

V0  =  11,500  —  o  =  11,500  pounds. 

F2  =  11,500  —  2000  =  9500  pounds. 

F4  =  11,500  —  4000  =  7500  pounds. 

F5'  =  11,500  —  5000  =  6500  pounds. 
F6"  =  11,500  —  5000  —  7000  =  —  500  pounds. 

V6=  11,500  —  6000  —  7000  =  — 1500  pounds. 

The  values  at  the  other  sections  were  obtained  in  the  same  way. 
The  shear  just  to  the  left  of  the  concentrated  load  is  different  from 
that  just  to  the  right;  consequently  the  shear  at  both  sections  must 
be  calculated. 

The  bending  moment  is  obtained  by  taking  the  moment  of 
the  reaction  about  an  axis  in  the  section  and  subtracting  from  it 
the  moment  of  the  loads  to  the  left  of  the  section  about  the  same 
axis. 

M0=  O, 

M2=  1 1, 500 X  24  —  2000 X  12  =  252,000  pound-inches, 
M4=  1 1, 500 X  48  —  4000  X  24  =  456,000  pound-inches, 
Af6=  1 1, 500 X  60  —  5000 X  30  =  540,000  pound-inches, 
M6=  1 1, 500 X  72  —6000X36  —  7000X12  =  528,000 pound-inches. 

The  bending  moments  at  the  other  sections  were  obtained  in 
the  same  way.  The  shear  passes  through  zero  at  5  feet  from  the 
left  support;  therefore  the  maximum  moment  occurs  at  that 
point. 


EXAM.]      BEAMS  —  EXTERNAL   FLEXURAL   FORCES  79 


In  Fig.  45  are  drawn  the  load,  shear,  and  moment  diagrams  for 
this  beam. 

3.  In  Example  No.  2  obtain  the  bending  moment  at  sections 
5  feet  and  8  feet  from  the  left  support  by  use  of  the  shear  diagram. 


X 

feet. 

V 

pounds. 

M 
pound- 
inches. 

0 

11,500 

0 

2 

9.50° 

252,000 

4 

7-500 

456,000 

5' 

6,500 

540,000 

5" 

-     500 

540,000 

6 

—  1,500 

528,000 

8 

-3.500 

468,000 

10 

-5.500 

360,000 

12 

-7.500 

204,000 

14 

-9.500 

0 

FIG.  45- 

Let  the  first  section  be  taken  under  the  load;  then  the  base  of 
the  trapezoid  is  5  X  12  =  60  inches;  the  average  ordinate  is 
(11,500  +  6500)  -*-  2  =  9000  pounds.  The  area  under  the  curve  is 
9000  X  60  =  540,000  pound-inches  =  M 6.  At  8  feet  the  negative 
area  must  be  taken  from  the  positive.  As  obtained  above,  the 
positive  area  is  540,000  pound-inches.  The  base  of  the  negative  area 
is  3  X  12  =  36  inches.  The  average  ordinate  is  (—  500  —  3500) 
-T-  2  =  —  2000  pounds,  therefore,  the  area  is  36  X  ( —  2000)  = 
—  72,000  pound-inches. 

M$=  540,000  —  72,000  =  468,000  pound-inches. 


8o      BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

PROBLEMS 

1.  A   simple  beam  of  i2-ft.  span  carries  a  uniform   load   of 
6000  pounds,  a  concentrated  load  of  3000  pounds  4  feet  from  the 
left  support,  and  one  of  6000  pounds  9  feet  from  the  left  support. 
Calculate  the  reactions.  Ans.   RI  =  6500  Ib. 

2.  A   24-ft.  beam   overhangs  the   right   support  4   feet.    It 
carries  a  uniform  load  of  900  pounds  per  foot  of  length  in  addition 
to  a  concentrated  load  of  10,000  pounds  9  feet  from  the  left  sup- 
port and  one  of  4000  pounds  at  the  right  end.     Determine  the 
reactions. 

3.  Calculate  the  vertical  shear  at  points  2  feet  apart  for  the 
beams  in  Problems  No.  i  and  2.     Plot  the  shear  diagrams. 

4.  Calculate  the  bending  moment  at  points  2  feet  apart  for 
the  beams  in  Problems  No.  i  and  2.     Plot  the  moment  diagrams. 

5.  If  a  simple  rectangular  timber  beam  of   i4-ft.  span  and 
depth  12  inches  carries  a  uniform  load  of  1000  pounds  per  foot 
and  is  cut  in  two  4  feet  from  the  left  support,  what  vertical  force, 
and  what  two_horizontal  forces,  if  8  inches  apart,  must  act  on  the 
left  portion  of  the  beam  at  the  cut  section  in  order  to  replace  the 
stresses  acting  between  the  two  portions  before  the  beam  was  cut  ? 

Ans.    V  =  —  3000  Ib.    H  =  30,000  Ib. 

For  the  following  beams  locate  the  danger  section,  calculate  max- 
imum bending  moment,  and  draw  the  load,  shear,  and  moment  dia- 
grams. Determine  the  bending  moment  by  getting  the  area  in  the 
shear  diagram. 

6.  A  i5-ft.  simple  beam  having  a  uniform  load  of  400  pounds 
per  foot  and  concentrated  loads  of  6000  pounds  and  5000  pounds 
at  5  feet  and  8  feet  respectively  from  the  left  support. 

7.  A  2o-ft.  simple  beam  having  concentrated  loads  of  2000 
pounds,  4000  pounds,  and  3000  pounds  at  4  feet,  6  feet,  and  12 
feet  respectively  from  the  left  support. 

8.  Simple  beam  of   2o-ft.  span  carrying  a  uniform  load  of 
120  pounds  per  foot  and  concentrated  loads  of  400  pounds,  600 
pounds,  and  600  pounds  at  4  feet,  6  feet,  and  16  feet  respectively 
from  the  left  support. 

9.  Cantilever  beam  of  8-ft.  span  with  a  load  of  10,000  pounds 
at  the  end.  Ans.  Mm  =  960,000  Ib.-in. 


PROB.]      BEAMS  —  EXTERNAL   FLEXURAL   FORCES  8 1 

10.  Cantilever  beam  of  zo-ft.  span  with  a  uniform  load  of 
15,000  pounds.  Ans.   Mm  =  900,000  Ib.-in. 

11.  Simple  beam  of  i2-ft.  span  with  a  concentrated  load  of 
9000  pounds  at  the  middle.  Ans.  Mm  =  324,000  Ib.-in. 

12.  Simple    beam    of    i5-ft.    span  with  a   uniform  load    of 
15,000  pounds.  Ans.   Mm  =  337,500  Ib.-in. 

13.  Simple  beam  of  i4-ft.  span  with  a  concentrated  load  of 
12,000  pounds  4  feet  from  left  support. 

14.  Simple  beam  of  40-ft.  span  with  a  uniform  load  of  20  tons, 
concentrated  load  of  100  tons  at  the  center. 

15.  Simple    beam  of    i6-ft.   span  with    a  uniform   load    of 
32,000  pounds,  and   a  concentrated   load  of   16,000  pounds  at 
4  feet  from  left  support.  Ans.  Mm  =  1,200,000  Ib.-in. 

1 6.  Simple  beam  of  i8-ft.  span  with  a  concentrated  load  of 
9000  pounds  4  feet  from  left  support,  one  of   7000  pounds  8 
feet  from  left  support,  one  of  12,000  pounds  13  feet  from  left 
support. 

17.  Simple  beam  of  i2-ft.  span  with  a  uniform  load  of  18,000 
pounds,  two  equal  concentrated  loads  of  10,000  pounds  each  at  the 
one-third  points. 

1 8.  Overhanging   beam    18    feet   long   overhanging    the   left 
support  4  feet,  with  a  uniform  load  of  1500  pounds  per  foot. 

19.  Simple  beam,  same  as  in  Problem  No.  14,  with  an  additional 
load  of  5000  pounds  at  the  center  of  the  beam. 

20.  A  simple  beam  of   2o-ft.   span  weighing   12  pounds  per 
lineal  foot,  with  a  load  of  240  pounds  5  feet  from  the  left  end. 

21.  A  simple  beam  of  20-ft.  span  with  concentrated  loads  of 
2000  pounds  4  feet  from  the  left  end,  and  1000  pounds  18  feet  from 
the  left  end,  and  also  a  uniform  load  of  100  pounds  per  lineal 
foot. 

22.  An  overhanging  beam  12  feet  long  overhanging  the  right 
end  2  feet  carrying  a  uniform  load  of  2000  pounds  per  foot  on 
entire  beam. 

23.  An  overhanging  beam  13  feet  long  overhanging  the  right 
support  3  feet,  carrying  a  uniform  load  of  1000  pounds  per  foot 
between  supports,  and  a  uniform  load  of  500  pounds  per  foot  on 
the  overhanging  end. 


82      BEAMS  —  EXTERNAL   FLEXURAL   FORCES      [CHAP.  V 

24.  Two  loads  each  of  3000  pounds,  5  feet  apart,  roll  over  a 
simple  beam  of  15 -ft.  span.     Find  the  position  of  these  loads 
for  the  maximum  bending  moment  and  find  its  value. 

Ans.   6|  ft.  from  left  support. 

25.  Two  wagon  wheels,  8  feet  apart,  roll  over  a  simple  beam  of 
24-ft.  span.     If  the  load  on  each  wheel  is  2000  pounds,  find  their 
position  for  the  maximum  bending  moment  and  determine  its 
value.  Ans.  Mm  =  200,000  Ib.-in. 

26.  Compute  the  maximum  bending  moment  due  to  two  loads 
of  1000  pounds  each  and  5  feet  apart  rolling  over  a  25-ft.  simple 
span. 

27.  Determine  the  maximum  bending  moment  produced  in 
a  beam  of  24-ft.  span  by  a  system  of  three  rolling  loads  of  weight 
10,000  pounds,  5000  pounds,  and  12,000  pounds  if  the  distance 
between  the  first  and  second  is  6  feet  and  the  distance  between 
the  second  and  third  is  5  feet. 


CHAPTER   VI 
BEAMS  — INTERNAL  FLEXURAL  STRESSES 

62.  FORCES  AND  STRESSES.  In  Chapter  V  the  ex- 
ternal forces  acting  on  beams  were  considered.  These 
external  forces  tend  to  cause  the  beam  to  rupture,  and 
that  tendency  is  resisted  by  the  internal  stresses  set  up 
in  the  beam.  The  nature,  distribution,  and  magnitude 
of  these  stresses  will  be  considered  in  this  chapter. 


' 

B 

C 

D 

E 

(a) 


A 

B 

D 

E 

C 

i 

F 

f  i.\ 

FIG.  46. 

The  vertical  shear  tends  to  rupture  the  beam  along  a 
vertical  plane  by  causing  one  portion  to  slip  past  the 
other.  To  illustrate  this  effect  imagine  a  beam  built 
up  of  several  blocks  glued  together  as  indicated  in 
Fig.  46  (a).  If  the  glue  becomes  soft  and  sticky,  the 
inner  portions  will  slide  down  past  the  others.  Or 
otherwise  stated  the  outer  portions  will  slide  up  past 
the  inner  portions.  The  magnitude  of  the  tendency  of 

83 


84      BEAMS— INTERNAL  FLEXURAL  STRESSES     [CHAP.  VI 

one  portion  to  slip  past  the  one  to  the  right  is  measured 
by  the  vertical  shear  at  the  plane  between  the  two  por- 
tions. This  vertical  shear  is  resisted  by  internal  resisting 
stresses  acting  vertically  at  the  section  considered. 

The  bending  moment  at  a  section  tends  to  cause  the 
end  of  the  beam  to  rotate  about  an  axis  in  that  section. 
Thus,  at  the  section  between  the  portions  A  and  B  of 
the  beam  in  Fig.  47  the  external  forces  tend  to  cause  A 


FIG.  47 

to  rotate  clockwise  about  an  axis  in  the  plane  between 
the  two  portions.  The  magnitude  of  this  tendency  is 
measured  by  the  bending  moment  at  the  section.  This 
is  resisted  by  the  moment  of  the  internal  stresses  which 
act  on  the  section  considered. 

In  the  common  theory  of  flexure  the  internal  resist- 
ing stresses  are  divided  into  their  vertical  and  hori- 
zontal components.  The  vertical  components  resist  the 
vertical  shear  and  the  moment  of  the  horizontal  com- 
ponents resist  the  bending  moment. 

63.  RESISTING  SHEAR.    THE  SHEAR  FORMULA.    As 

already  defined  the  vertical  shear  V  at  a  section  is  the 
algebraic  sum  of  the  external  forces  acting  on  the  por- 
tion of  the  beam  to  the  left  of  the  section;  and  this  ver- 
tical shear  is  resisted  by  the  stresses  in  the  fibers  between 
the  two  portions  of  the  beam.  This  stress,  which  re- 
sists the  vertical  shear,  represented  in  Fig.  48  by  V, 
is  called  the  resisting  shear.  If  the  maximum  shearing 
unit-stress  in  the  cross  section  is  s  the  average  value 
will  be  ks,  where  k  is  a  constant  depending  upon  the 


ART.  64] 


THE    SHEAR   FORMULA 


shape  of  the  cross  section,  and  if  A  is  the  area  of  the 
cross  section  the  resisting  shear  equals  ksA.  Therefore, 
since  equilibrium  exists,  the  vertical  shear  equals  the 
resisting  shear  and  the  shear  formula  is 

V=ksA.  (i) 


w,         w0        w, 

LI" 


FIG.  48. 


64.  THE  VALUE  OF  &  IN  THE  SHEAR  FORMULA.    As 

shown  in  Chapter  XIII,  the  intensity  of  the  shearing 
unit-stress  is  not  the  same  at  all  points  of    the  cross 
section.     The  maximum  stress  from  the  shear  formula 
V 


IS  S  = 


kA 


3  V 
For  a  rectangular  beam  s  =  -  -j , 

2      ^T. 


and   &  =  - 


For  a  circular  beam 


i  =  3l'   and 


4 


For  I-beams  and  built-up  sections  it  is  approximately 
assumed  that  the  maximum  stress  is  equal  to  that 
obtained  by  dividing  the  shear  by  the  area  of  the  web; 
then  the  maximum  stress  is- 


y 

s  =  -r  ,  where  A\  is  the  area  of  the  web. 


86     BEAMS— INTERNAL  FLEXURAL  STRESSES      [CHAP.  VI 


If  k  for  this  case  is  desired  it  is  equal  to  -j  .     The  assump- 

A. 

tion  is  only  approximately  correct  but  the  values  are  near 
enough  the  true  ones  to  be  used  in  design. 

65.  RESISTING  MOMENT.  If  a  beam  were  cut 
through  at  a  section,  as  at  AB,  Fig.  49,  and  the  same 
external  forces  were  to  continue  to  act  on  the  left  por- 
tion, besides  a  vertical  resisting  shear  ksA,  forces  equal 
to  the  horizontal  forces,  as  Hc  and  Ht,  should  be  supplied 
to  produce  equilibrium.  These  forces  are  the  hori- 


r   r  r 


FIG.  49. 

zontal  components  of  the  internal  stresses  acting  on 
the  given  section  of  the  beam  and  make  up  the  hori- 
zontal resisting  forces  which  produce  the  resisting 
moment. 

66.  ASSUMPTIONS  FOR  THE  RESISTING  MOMENT. 
The  moment  formula  to  be  derived  is  for  beams  made 
of  materials  that,  have  the  same  modulus  of  elasticity  in 
tension  and  compression.  The  formula  is  true  only 
for  stresses  less  than  the  elastic  limit  of  the  material. 
It  is  assumed  that  a  transverse  plane  section  of  the  beam 
before  bending  remains  a  plane  section  after  bending. 
From  these  assumptions  the  nature,  distribution,  and 
magnitude  of  the  stresses  producing  Hc  and  Ht  may  be 
found. 


ART.  67]      DISTRIBUTION   OF   THE   FIBER   STRESSES        87 

67.  DISTRIBUTION  OF  THE  FIBER  STRESSES.  Two 
sections  of  the  beam,  as  AB  and  CD  in  Fig.  50  (a), 
parallel  before  the  beam  is  bent,  assume  positions  shown 
in  Fig.  50  (&)  after  bending  occurs.  All  fibers  of  the 
beam,  except  those  along  the  surface  OX,  will  be  length- 
ened or  shortened,  thus  having  stresses  developed  in 
them.  The  surface  OX  along  which  no  tensile  or  com- 
pressive  stresses  are  developed  is  the  neutral  surface. 
Its  intersection  //  with  a  transverse  section  is  the 


1 

1 

, 

EB 

F 

(a) 


'Fie.  50. 


neutral  axis.  Since  the  modulus  of  elasticity  E  is 
considered  constant,  all  other  fibers  will  have  a  stress 
proportional  to  their  deformation.  Take  the  fiber  EF 
at  a  distance  y  from  the  neutral  surface.  Draw  C'D' 
parallel  to  AB  in  Fig.  50  (b)  ;  then  ey  represents  the  de- 
formation of  that  fiber  in  the  length  EF  after  bending. 
The  deformation  ey  divided  by  the  length  EF  gives  the 
unit  deformation.  If  this  be  multiplied  by  the  modulus 
of  elasticity  of  the  material  the  result  will  give  the 
unit-stress  fy  coming  on  the  fiber.  The  direction  of 
the  stress  is  the  same  as  that  of  the  deformation  which 
is  along  the  line  EF  and  acts  normal  to  the  section. 


88      BEAMS— INTERNAL  FLEXURAL  STRESSES      [CHAP.  VI 

Since  the  greatest  deformation  occurs  in  the  fiber 
farthest  from  the  neutral  axis  the  greatest  stress  will 
be  in  that  fiber. 

It  is  seen  that  the  deformation  of  a  fiber  is  propor- 
tional to  its  distance  from  the  neutral  surface,  conse- 
quently the  stress  is  proportional  to  the  distance  of  the 
fiber  from  the  neutral  axis.  The  ratio  of  the  maximum 
unit-stress  in  compression  to  the  maximum  unit-stress 
in  tension  is  the  same  as  the  ratio  of  the  distance  of  the 
most  remote  fiber  in  compression  from  the  neutral  axis 
to  the  distance  of  the  most  remote  fiber  in  tension 
from  the  neutral  axis,  hence 

jrl: 

where  fe  and  ft  are  the  maximum  stresses  in  compression 
and  tension,  and  cc  and  ct  are  the  distances  from  the 
neutral  axis  to  the  most  remote  fibers  in  compression 
and  tension  respectively. 

Fig.  51  (a)  shows  a  free- body  diagram  of  the  left 
portion  of  a  beam,  in  which  the  resisting  stresses  in 
tension  and  compression  are  indicated  in  intensity  by 
the  length  of  the  vectors  representing  them.  Fig.  51  (b) 
is  an  end  view  of  the  section,  in  which  the  stresses  acting 
normal  to  the  section  (also  normal  to  the  plane  of  the 
paper)  are  indicated  by  crosses  for  compression  and  by 
circles  and  dots  for  tension.  The  intensity  of  the  stress 
is  indicated  by  the  weight  of  the  lines. 

68.  POSITION  OF  THE  NEUTRAL  SURFACE  AND  THE 

NEUTRAL  AXIS.  Let  Fig.  51  (a)  represent  a  portion  of 
a  beam  under  load  and  Fig.  51  (b)  represent  the  cross 
section.  Let  the  maximum  fiber  stress  developed  be  /, 
which  comes  on  the  fiber  most  remote  from  the  neutral 
axis  //.  Call  the  distance  of  this  fiber  from  the  neutral 


ART.  68]      NEUTRAL   SURFACE   AND   NEUTRAL  AXIS        89 

axis  c,  then  the  unit- stress  developed  on  a  fiber  at  the 
distance  y  is  -/,  and   the  total  stress  acting  upon  the 

small  area  a  is  ^/a.     The  sum  of  the  horizontal  stresses 

T2 


I 


/CSA^ 
iff 


RI 


X     I 


X* 

xxxvx 


0  O  O  O 

©©00® 


RI 


FIG.  51. 


acting  upon  the  entire  cross  section  is  equal  to  the  alge- 
braic sum  of  all  stresses  on  the  elements  of  areas.  Since 
these  are  the  only  horizontal  forces  acting  upon  the  left 
portion  of  the  beam  this  sum  is  equal  to  zero. 


y  =  o. 

2  ay  is  the  moment  of  the  area  with  respect  to  the 
neutral  axis  and  y  is  the  distance  of  the  centroid  of 
the  cross  section  from  the  neutral  axis.  Therefore,  the 
neutral  surface  passes  through  the  centroid  of  the  cross 
section  of  a  beam. 


90      BEAMS— INTERNAL  FLEXURAL  STRESSES    [CHAP.  VI 

69.  THE  MOMENT  FORMULA.    In  Fig.  51  let  c  be 

the  distance  of  the  most  remote  fiber  from  the  neutral 
axis,  /  the  tensile  or  compressive  unit-stress  developed 
on  that  fiber.  The  unit-stress  on  a  fiber  at  a  distance  y 
from  the  neutral  axis  is 

,       *-** 

The  total  stress  coming  on  a  small  area  a  is  a  -/.  The 
moment  of  this  stress  about  the  neutral  axis  is 

a^-fy  =  -ay*. 
cj'       c   • 

To  get  the  moment  of  all  the  stresses  developed  in  the 
section  all  such  expressions  must  be  summed  up,  giving 
the  resisting  moment  equal  to 

2^  a/   or   i^ay*. 
c  c 

The  expression  2  ay*  is  called  the  moment  of  inertia  of 
the  cross  section  for  the  neutral  axis  and  is  denoted  by  /. 
(See  Appendix  A.)  By  substituting  /  for  2  ay*  the  re- 
sisting moment  becomes  — »  and  as  equilibrium  exists 

the  bending  moment  M  equals  the  resisting  moment  i* . 
therefore,  -,. 

M  =  }-j-  (2) 

In  the  moment  formula  /  is  the  maximum  tensile  or 
compressive  unit-stress  existing  in  the  section  for 
which  the  bending  moment  is  M.  This  stress  is  developed 
in  the  fiber  most  remote  from  the  neutral  axis,  which 
is  at  the  distance  c  from  that  axis.  A  stress  obtained 
by  the  use  of  the  moment  formula  is  called  a  fiber  stress. 

The  quantity  -  depends  upon  the  size  and  shape  of  the 
c 

cross  section  of  the  beam  and  is  called  the  section  factor 
or  the  section  modulus. 


ART.  71]      COMPRESSIVE   AND   TENSILE   STRESSES  QI 

70.    UNITS.     The  unit  of  /  depends  upon  those  of  M 

and  -  •     If  M  be  in  ton-inches,  /  is  in  tons  per  square 

inch.  If  M  is  in  pound-inches,  /  is  in  pounds  per  square 
inch.  The  unit  of  5  depends  upon  those  of  V  and  A. 
The  units  used  will  be  pounds  per  square  inch  for  stresses, 
pounds  for  loads,  square  inches  for  areas,  pound-inches 
for  moments,  (inches)4  for  moments  of  inertia,  (inches)3 
for  section  factors,  and  inches  for  distances. 


ILLUSTRATIVE  EXAMPLE 

As  an  illustration  of  the  application  of  this  formula  let  it  be 
required  to  determine  the  maximum  fiber  stress  developed  at  a 
certain  section  of  a  beam  where  M  =  115,200  pound-inches  as 
calculated  by  the  principles  of  Chapter  V.  /  =  21.8  inches4  and 
c  =  3  inches  as  calculated  by  the  methods  of  Appendix  A.  Then 

,      Me      115.200  X  3 
/  -  —  -  -     2i8     -  =  15,850  Ib.  per  sq.  in. 

71.  TOTAL  HORIZONTAL  COMPRESSIVE  AND  TENSILE 

STRESSES.     From  Art.  68,  the  stress  on  an  element  of 
area  a  at  the  distance  y  from  the  neutral  axis  is 


To  obtain  Hc,  which  is  the  resultant  of  the  compressive 
stresses,  a  summation  of  the  stresses  on  the  area  in  com- 
pression must  be  made.  Therefore, 

Hc  =  -2ay, 
Hc=S-y'A', 

0 

in  which  y'  is  the  distance  from  the  neutral  axis  to 
the  centroid  of  the  area  A'  which  is  in  compression. 


92      BEAMS—  INTERNAL  FLEXURAL  STRESSES      [CHAP.  VI 

A  similar  expression  is  obtained  for  the  resultant  of  the 
tensile  stresses  which  is 


If  the  bending  moment  is  positive  the  compression  area 
is  above  the  neutral  axis.     In  the  formula 


Hc  =    y'A', 

-  y'  is  the  unit-  stress  on  a  fiber  at  the  centroid  of  the  area 
c 

A'  .  Therefore,  the  resultant  of  the  horizontal  compres- 
sive  stresses  equals  the  stress  on  the  centroid  of  the  area 
in  compression  multiplied  by  the  area  in  compression. 
There  is  a  similar  principle  for  tension.  This  principle 
is  general  and  can  be  applied  to  any  portion  of  a  cross 
section  of  a  beam  when  the  stresses  are  below  the 
elastic  limit. 

The  line  of  action  of  the  resultant  of  the  compressive 
or  of  the  tensile  stresses  is  obtained  by  the  principle  of 
moments  for  a  system  of  forces,  which  is,  the  moment 


T 

W2fcs 

A' 

\ 
I 

f 

P£ 

X 

» 

^ 

J 

FIG.  52. 

of  the  resultant  of  a  system  of  forces  about  any  axis 
is  equal  to  the  summation  of  the  moments  of  all  the 
forces  about  the  same  axis.  Let  dr  be  the  distance 


ART.  72]  THE  THREE  PROBLEMS  93 

from  the  neutral  axis  to  the  resultant  of  the  compressive 
stresses,  then  (see  Fig.  52), 


~  y'A'     G' 

as  Hc  =  -y'd'j  and  I'  is  the  moment  of  inertia  of  the 

area  in  compression  about  the  neutral  axis,  and  G'  is 
the  moment  of  that  area  about  the  same  axis.  A 
similar  expression  is  obtained  for  the  distance  to  the 
center  of  tension. 

As  Hc  and  Ht  are  equal  they  produce  a  couple.  The 
arm  between  the  center  of  compression  and  the  center 
of  tension  multiplied  by  Hc  or  Ht  equals  the  resisting 
moment. 

T/  7 

For  a  rectangular  beam  of  depth  d,  -^-,  =  -  •     The  dis- 

tance between  the  center  of  compression  and  the  center 
of  tension,  then,  is  f  d. 

72.  THE  THREE  PROBLEMS.    In  each  one  of  the  two 

fundamental  beam  formulas,  the  shear  formula  V=  ksA, 

and  the  moment  formula  M  =  —  >  there  are  three  vari- 

c  / 

ables:  V,  s,  and  A  in  the  former  and  M,  /,  and-  in  the 

G 

latter.  Any  one  of  the  three  variables  in  each  equation 
may  be  determined  if  the  other  two  are  known.  This 
gives  rise  to  three  problems  that  may  be  investigated 
by  the  use  of  the  shear  and  moment  formulas. 

PROBLEM  I.  Investigation  of  Beams.  Given  a  beam 
with  its  load,  to  calculate  the  maximum  unit-stresses. 


94      BEAMS  —  INTERNAL  FLEXURAL  STRESSES      [CHAR  VI 

By  the  principles  developed  in  Chapter  V  the  values 
of  the  maximum  vertical  shear  and  the  bending  moment 

may  be  calculated.     A  and  -  may  be  determined  from 

c 

the  dimensions  of  the  cross  section  of  the  beam,  by  the 
methods    of    Appendix   A.     To    obtain    the    maximum 

y 
shearing  stress,  s  =  -7— T-  may  be  used,  and  to  obtain  the 

,,,  ,      Me 

fiber  stress,/  =  -y-  may  be  used. 


ILLUSTRATIVE  EXAMPLE 

Calculate  the  maximum  shearing  stress  and  the  maximum 
fiber  stress  developed  in  a  longleaf  pine  beam  of  lo-ft.  span, 
breadth  4  inches,  depth  8  inches,  when  carrying  a  concentrated 
load  of  12,000  pounds  4  feet  from  the  left  support. 

Ri  =  720  pounds,     Vm  =  720  pounds, 
Sm  =  \  (720  -T-  32)  =  33. 7  pounds  per  square  inch, 
Mm  =  720  X  48  =  34, 560 pound-inches, 
^4X8X8X8^512. 

12  3 

c  =  4  inches, 
...  f  =  34,56o  X  4  X  3  =  glo  pounds  per  square  inch< 

PROBLEM  II.     Safe  Loads  for  Beams.     By  the  use  of 

the  shear  and  moment  formulas,  the  load  which  a  given 

beam  will  safely  carry  may  be  obtained.     M  =  —  gives 

c 

the  value  of  the  maximum  allowable  bending  moment, 
from  which  values  the  load  may  be  selected.  After  de- 
termining the  load  by  use  of  the  moment  formula,  the 
beam  should  be  investigated  for  the  maximum  shearing 
stress  developed  by  that  load  by  use  of  the  shear  formula 
V=  ksA.  The  allowable  shearing  stress  or  bending  mo- 
ment as  calculated  should  not  be  exceeded. 


ART.  72]  THE  THREE   PROBLEMS  95 

ILLUSTRATIVE  EXAMPLE 

What  uniform  load  will  a  lo-inch,  25-pound  I-beam  carry 
when  used  as  a  simple  beam  of  i6-ft.  span  with  an  allowable  fiber 
stress  of  16,000  pounds  per  square  inch?  From  Table  20  which 

gives  values  of  section  factors  of  I-sections,  -  =  24.4  inches3. 

c 

The  maximum  allowable  resisting  moment  is  16,000  X  24.4  =  390,400 
pound-inches.  The  maximum  bending  moment  for  the  uniformly 

Wl 
distributed  load  on  a  simple  beam  occurs  at  the  center  and  is  —  • 

o 

W  X  192 

-  =  390,400, 

o 

W  =  1 6, 2  70  pounds. 

To  get  the  approximate  shearing  stress,  the  area  of  the  web  is 
10  X  0.31  =  3.1  square  inches.  /.  s  =  (16,270  -5-2)^-  3.1  =  2630 
pounds  per  square  inch,  which  is  safe. 

PROBLEM  III.  The  Design  of  Beams.  The  loading 
of  a  beam  and  the  maximum  allowable  stress  may  be 
specified,  to  select  or  design  a  beam  to  carry  the  load 
safely.  The  design  of  beams  is  the  problem  most 
generally  met  with  by  the  engineer  and  architect.  It 
admits  of  many  solutions,  and  the  designer  must  use  his 
best  judgment  in  choosing  the  form  and  size  of  the  cross 
section  to  be  used.  The  material  most  remote  from  the 
neutral  axis  is  all  that  is  stressed  to  the  maximum,  while 
that  at  the  neutral  axis  has  no  fiber  stress  in  it.  The 
material  is  most  efficiently  used  when  the  largest  pro- 
portion of  it  is  stressed  nearly  to  the  maximum  stress, 
and  obviously  this  condition  exists  in  a  section  having 
a  large  part  of  the  material  well  away  from  the  neutral 
axis.  -Necessarily  there  must  also  be  such  a  distribu- 
tion of  the  material  as  will  insure  safety  against  shear, 
buckling,  and  twisting.  In  steel  I-beams  there  is  a 
large  portion  of  the  material  near  the  outside  fiber, 
and  yet  the  web  is  large  enough  to  resist  the  shear. 


96      BEAMS  — INTERNAL  FLEXURAL  STRESSES    [CHAP.  VI 

Generally,  rupture  is  due  to  bending  rather  than  to 
shear,  and  occurs  at  the  danger  section.  In  the  deter- 
mination of  the  safe  loads  for  beams  and  the  design  of 

beams,  the  moment  formula  M  =  —  is  the  governing 

c 
y 

formula,  the  shear  formula  5  =  T-J  being  used  after- 
ward to  investigate  the  beam  for  the  shearing  unit- 
stress.  If  the  shearing  unit-stress  developed  is  too  large 
another  design  must  be  made,  but  this  is  seldom  neces- 
sary, except  for  short  deep  beams. 

ILLUSTRATIVE  EXAMPLE 

Design  a  square  loblolly  pine  cantilever  beam  for  a  span  of  8  feet 
with  a  concentrated  load  of  500  pounds  at  the  free  end.  From 
the  table  of  allowable  stresses  (No.  8)  /  =  1000  pounds  per  square 
inch.  Mm  =  —  500  X  96  =  -  48,000  pound-inches.  (The  nega- 
tive sign  may  be  neglected,  as  that  simply  means  that  the  stress 
in  the  top  fibers  is  tension.)  If  b  is  the  breadth  of  the  section, 

I-  =  -, 
c       6' 

73 

.*.     48,000  =  1000  X  —  > 
6 

b3  =  288, 
b  =  6. 6 1  inches. 

The  maximum  shearing  unit-stress  for  this  load  is 
s  =  3jf  =  3  x  5oo  =          lb  . 

2  A          2         43.7 

which  is  safe.  While  the  above  beam  satisfies  the  condition  of 
the  problem  it  is  not  a  standard  section  and  probably  would  be 
replaced  by  a  6-inch  by  8-inch  beam,  for  which  case  the  fiber 
stress  is 

=  48,000X4X12  = 
6X8X8X8 

The  maximum  shearing  stress  is 

s  =  -  X  ^  =  15.6  lb.  per  sq.  in. 
2       48 


ART.  74] 


MAXIMUM   STRESS   DIAGRAMS 


97 


73.    MODULUS  OF   RUPTURE.     The  moment  formula 
M  =  —  is  not  applicable  to  beams  of  material  for  which 

the  stress  is  not  proportional  to  the  deformation,  or  for 
non-homogeneous  beams,  or  for  beams  under  stresses 
greater  than  the  elastic  limit  of  the  material.  However, 
it  is  frequently  used  to  calculate  a  nominal  unit-stress 
developed  in  a  beam  when  the  bending  moment  is  great 
enough  to  cause  failure.  The  unit- stress  thus  calculated 
is  called  the  modulus  of  rupture.  This  usually  lies 
between  the  ultimate  compressive  strength  and  the  ulti- 
mate tensile  strength  of  the  material.  Table  13  gives 
average  values  for  the  modulus  of  rupture. 


TABLE  13 

MODULUS  OF   RUPTURE 


Material. 

Modulus  of  rupture 
Ib.  per  sq.  in. 

Timber 

7000  to  oooo 

Cast  iron  

35.000 

74.  MAXIMUM  STRESS  DIAGRAMS.    The  value  of  the 

maximum  shearing  unit-stress  for  a  section  can  be 
obtained  by  dividing  the  vertical  shear  for  the  section 
by  kA,  the  product  of  the  sectional  area  and  a  constant 
k  depending  upon  the  shape  of  the  section.  The  value 
of  the  maximum  fiber  stress  for  the  section  can  be  ob- 
tained by  dividing  the  bending  moment  at  that  section 

by  -,  the  section  modulus.     If  the  shear  and  moment 
c 

diagrams  are  drawn,  the  values  of  V  and  M  may  be 
taken  directly  from  the  diagrams;  thus,  in  Fig.  53  (b), 
CD  divided  by  kA  (assumed  constant)  gives  the  shear- 
ing unit-stress.  The  stress  is  always  proportional  to 


98      BEAMS  —  INTERNAL  FLEXURAL  STRESSES     [CHAP.  VI 


the  ordinate  CD. 


in  Fig.  53  (c)  is  the  bending 


moment  at  the  section  which  divided  by   -  gives  the 

0 

maximum  fiber  stress,  which  comes  in  the  extreme  fiber 
of  the  beam  at  that  section.     The  fiber  stress  is  pro- 


(a) 


portional  to  the  ordinate  of  the  moment  curve.  Thus 
for  a  beam  of  uniform  section  the  ordinates  to  the 
vertical  shear  curve  represent  the  maximum  shearing 
stresses,  and  the  ordinates  to  the  bending  moment  curve 
represent  the  maximum  fiber  stresses  existing  in  the 
beam.  Consequently  the  shear  and  moment  diagrams 
may  be  considered  stress  diagrams.  The  above  reason- 
ing is  for  beams  of  constant  section,  but  with  modifica- 
tions similar  reasoning  may  be  applied  to  beams  in  which 
the  section  is  not  uniform. 


EXAM.]      BEAMS  —  INTERNAL  FLEXURAL   STRESSES       99 

•75.  BEAMS  OF  UNIFORM  STRENGTH.    If  a  beam  is 

of  uniform  section  the  maximum  fiber  stress  occurs  only 
in  the  outside  fiber  at  the  section  of  the  greatest  bending 
moment.  The  stress  varies  with  the  bending  moment 
along  the  length  of  the  beam.  In  order  to  have  the 
most  efficient  beam  all  the  material  in  it  should  be 
stressed  to  the  allowable  stress,  and  to  approach  this 
state,  besides  keeping  the  material  near  the  outside 
surface,  the  cross  section  is  sometimes  made  to  vary 
with  the  bending  moment.  This  is  done  in  plate  girders 
where  extra  cover  plates  are  added  toward  the  center 
of  the  span. 


EXAMPLES 

i.  If  a  4-inch  by  6-inch  by  o. 4-inch  channel  is  used  as  a  simple 
beam  of  8-ft.  span  with  a  concentrated  load  of  2000  pounds 
three  feet  from  the  left  support,  (a)  what  is  the  maximum  fiber 
stress  developed?  (b)  What  is  the  stress  developed  on  a  fiber 
2  inches  from  the  top  and  2  feet  from  the  right  support? 

RI  =  1 2  50  pounds. 

(a)  The  maximum  moment  is  under  the  load  and  is 

Mm  =  1250X3X12=  45,000  pound-inches. 

The  centroid  is  1.29  inches  from  the  back  (see  Fig.  A6,  Appen- 
dix A).  Therefore 

c  =  4  —  1.29  =  2.71  inches, 
A  =5.28  square  inches, 
/  =  8.41  inches4, 

,      Me      45,000  X  2.71 
f  =:  __  =  ±  ___ =  I4?soo  lb  per  sq  m> 

(b)  Mn  =  1250  X -8  X  12  -  2000  x  3  X  12  =  18,000 Ib.-in. 

y  =  2.71  —  2  =  .71  inch. 

f        18.000  X  0.71 

fv  = — —        =  1520  lb.  per  sq.  in. 


100     BEAMS— INTERNAL  FLEXURAL  STRESSES    [CHAP.  VI 

2.  What  uniform  load  will  a  4-inch  by  6-inch  yellow  pine 
timber  safely  carry  when  used  as  a  simple  beam  of  lo-ft.  span? 

w; 

/       12        bd?       1X6x6 

-C  =  T  =  T  =  -— - 

2 

The  allowable  stress  is  1000  pounds  per  square  inch.     (Table  8), 

—  =  looo  X  24  =  24,000  pound-inches, 
c 

which  is  the  allowable  resisting  moment. 
The  maximum  bending  moment  is 

Wl      W  X  10  X  12 

—  =  -    — —     -  =  24,000, 

o  o 

W  =  1600  pounds. 

The  load  per  lineal  foot  is  ^ §a  =  160  pounds  per  foot.  Inves- 
tigating for  the  shearing  stress,  s  =  f  X  -V?0  =  5°  pounds  per 
square  inch. 

3.  Using  the  results  of  Example  No.  2,  what  is  the  total  com- 
pressive  stress  at  a  section  30  inches  from  the  left  end,  and  where  is 
the  line  of  action  of  the  resultant  ? 

Ri  =  800  pounds. 
MM  =  800  X  30  -  -W  X  30  X  15  =  18,000  Ib.-in. 

The  area  in  compression  is  12  square  inches.  The  stress  on 
the  centroid  of  this  area  is 

,        18,000  X  i.<5 
fv  = — —   -  =  375  ib.  per  sq.  in. 

The  total  compressive  stress  is  12X375  =  4500  pounds. 
The  line  of  action  of  the  resultant  is  I  the  depth  from  the  top  or 
i  inch  from  the  top  or  2  inches  above  the  neutral  surface. 

4.  Design  a  simple  cypress  timber  beam  to  carry  a  uniform 
load  of  8000  pounds  on  a  span  of  1 2  feet. 

^.w/.  8000x144. 

o  o 


PROB.]      BEAMS  —  INTERNAL   FLEXUfcAE-  &TR2SSE& \  ^M 

The  beam  must  be  large  enough  to  take  this  moment  without 
exceeding  the  allowable  stress  which  is  1000  pounds  per  square 
inch.  (Table  8), 

M       144,000  .    ,     ,      7 

_  =  _         _  =  IAA  inches3  =  -  • 

/  1000  C 

An  indefinite  number  of  cross  sections  will  satisfy  this,  but  the 
one  chosen  should  be  economical  and  suited  for  the  purpose.  It 
must  be  wide  enough  to  prevent  lateral  bending.  If  6  is  assumed 

equal  to  -  then 

7       bd?      d* 

-  =  —  =  —  =  144,  <f  =  1728,  a  =  12  inches,  and  0  =  6  inches. 

C  O  12 

The  maximum  shearing  stress  is 

^  =  |  X  4?i°  =  83!  Ib.  per  sq.  in., 

which  is  safe.  Other  dimensions  to  the  nearest  even  inch  above 
the  actual  size  required  are  2  inches  by  18  inches;  4  inches  by 
16  inches;  10  inches  by  10  inches;  or  three,  3  inches  by  10  inches. 
The  beam  2  inches  by  18  inches  would  be  too  narrow  and  deep 
unless  well  braced  laterally.  The  one  4  inches  by  16  inches  might 
be  chosen  in  some  cases  if  well  braced  laterally.  The  one  10  inches 
by  10  inches  would  not  be  economical  unless  the  two  inches  verti- 
cal distance  saved  would  be  more  valuable  than  the  extra  mate- 
rial in  the  beam  of  this  size. 


PROBLEMS 

i.  If  a  i5-inch,  42-pound  I-beam  carries  a  uniformly  dis- 
tributed load  of  39,270  pounds  on  a  span  of  16  feet,  (a)  What  is 
the  maximum  shearing  stress?  (b)  Draw  the  load,  shear,  and 
moment  diagrams,  (c)  Calculate  the  maximum  fiber  stress, 
(d)  What  is  the  resultant  of  the  horizontal  compressive  or  tensile 


10?     BEAMS-  —  INTERNAL  FLEXURAL  STRESSES    [CHAP.  VI 

stresses  at  the  danger  section?  (e)  What  is  the  rate  of  change 
of  the  vertical  shear  at  any  section?  (f)  What  is  the  rate  of 
change  of  the  bending  moment  at  the  left  support,  at  the  quarter 
point,  and  at  the  center  of  the  span  ? 

Ans.    (c)  16,000  Ib.  per  sq.  in. 

2.  (a)  What  uniform  load  will  a  simple  rectangular  Washing- 
ton fir  beam  of  breadth  8  inches  and  depth  12  inches  carry  on  a 
span  of  12  feet?     (b)  With  the  calculated  load  on  the  beam  what 
is  the  maximum  shearing  stress?     (c)  What  is  the  value  of  the 
fiber  stress  4  inches  from  the  top  of  the  beam  and  4  feet  from  the 
right  support  ?     (d)  What  is  the  maximum  shearing  stress  on  that 
section?     (e)  What  is   the  resultant  of   the    horizontal   tensile 
stresses  at  that  section?     (f)  Where  is  the  line  of  action  of  the 
resultant  of  the  tensile  stresses  ? 

3.  In  a  table  giving  the  safe  load  in  pounds  uniformly  distributed 
for  rectangular  beams  of  white  pine,  cedar,  and  spruce  for  each 
inch  in  thickness  the  following  values  are  given:  span  10  feet, 
depth  14  inches,  load  1524  pounds;  span  16  feet,  depth  21  inches, 
load  2144  pounds;  span  25  feet,  depth  22  inches,  load  1255  pounds. 
What  unit-stress  was  allowed  in  compiling  that  table  ? 

Ans.    700  Ib.  per  sq.  in. 

4.  With  an  allowable  unit-stress  of  noo  pounds  per  square 
inch  what  will  be  the  allowable  uniform  loads  per  inch  thickness 
of  the  beam  for  the  following  spans  and  depths?     Span  8  feet, 
depth  12  inches;  span  n  feet,  depth  14  inches;   span    20   feet, 
depth  24  inches. 

5.  Solve  Problem  No.  4  if  the  allowable  stress  is  1200  pounds 
per  square  inch.  Ans.     2376  Ib. 

6.  A  i2-inch,  40-pound  I-beam  of  a  span  of  20  feet  is  used  to 
carry  a  uniform  load  of  500  pounds  per  foot  and  a  concentrated 
load  of  5000  pounds  4  feet  from  the  left  end.     What  is  the  maxi- 
mum stress  developed? 

7.  Compute  the  maximum  unit-stress  in  a  2  X  8  inch  joist 
carrying  loads  of  240  pounds  3  feet  from   the  left  end  and  of 
360  pounds  4  feet  from  the  right  end  of  a  simple  span  of  12 
feet. 

8.  Determine  maximum  fiber  stress  in  a  6-inch  by  1 2-inch 
simple  beam  of   i2-ft.  span  which  carries  a  uniform  load  of 


PROB.]      BEAMS  —  INTERNAL  FLEXURAL  STRESSES      103 

100  pounds  per  foot  and  three  concentrated  loads  of  1300  pounds, 
1500  pounds,  and  1000  pounds  at  3  feet,  5  feet,  and  8  feet  respec- 
tively from  the  left  support. 

9.  Compute  the  maximum  fiber  stress  in  a  1 5-inch,  42-pound 
I-beam,  over  a  simple  span  of  30  feet  carrying  a  uniform  load  of 
500  pounds  per  foot  and  two  concentrated  loads  of  5000  pounds 
and  10,000  pounds  at  3  feet  and  23  feet  respectively  from  the  left 
support. 

io!  A  rectangular,  cantilever  timber  beam  of  i2-ft.  span, 
4  inches  broad,  and  8  inches  deep  carries  a  uniform  load  of  50 
pounds  per  lineal  foot.  Find  the  maximum  fiber  stress. 

11.  A  cantilever  white  cedar  beam  of  5-ft.  span  has  a  rectan- 
gular section  2  inches  broad  and  3  inches  deep.     Find  the  total 
uniform  load  it  can  safely  carry. 

12.  Find  the  uniform  load  per  lineal  foot  which  a  wooden 
cantilever  beam  6  feet  in  length,  of  rectangular  section  2  inches 
broad  and  3  inches  deep,  can  carry  with  a  maximum  fiber  stress  of 
800  pounds  per  square  inch. 

13.  A  i5-inch,  4  2 -pound  I-beam  is  carrying  a  total  uniform 
load  of  30,000  pounds  on  a  simple  span  of  20  feet.     Compute 
the  intensity  of  stress  at  a  point  3  inches  below  the  top  flange 
face  and  6  feet  from  the  left  support. 

14.  Determine   the   total  amount  of  horizontal   compressive 
stress  at  the  section  of  maximum  bending  moment  in  a  6-inch  by 
1 2-inch  wooden  beam  carrying  a  uniform  load  of  4000  pounds  per 
foot  on  a  simple  span  of  1 2  feet. 

15.  Design   a  cypress  beam  of  i8-ft.  span  to  carry  a  load 
that  varies  uniformly  from  zero  at  the  ends  to  a  maximum  of 
1800  pounds  per  foot  at  the  center. 

16.  Select  the  proper  I-beam  18  feet  long  which  overhangs 
both  supports  3  feet  that  will  carry  concentrated  loads  of  5000 
pounds  at  the  left  end,  10,000  pounds  at  the  center,  and  8000 
pounds  at  the  right  end. 

17.  Determine  the  maximum  fiber  stress  at  the  sections  indi- 
cated in  an  8-inch  by  1 2-inch  simple  beam  of  i2-ft.  span  which 
carries  a  uniform  load  of  800  pounds  per  foot:  (a)  2  feet  from  the 
supports,  (b)  at  the  quarter  points,  (c)  at  the  center. 

Ans,  900"  pounds  per  square  inch. 


104     BEAMS— INTERNAL  FLEXURAL  STRESSES   [CHAP.  VI 

18.  A  6-inch  by  1 2-inch  cantilever  beam  of  g-ft.  span  carries  a 
load  of  500  pounds  per  foot.     Calculate  the  fiber  stress  (a)  2  inches 
from  the  top  at  the  wall,  (b)  on  the  bottom  fiber  at  the  wall, 
(c)  on  the  top  fiber  at  the  middle.       Ans.    1687  Ib.  per  sq.  in. 

19.  A  2-inch  by  4-inch  maple  tirrfber  10  feet  long  is  to  be  used 
as  a  simple  beam.    What  central  load  will  it  safely  carry  (a)  when 
it  is  laid  flat,  and  (b)  when  placed  on  edge  ?    What  is  to  be  learned 
from  the  results? 

20.  What  should  be  the  depth  of  a  rectangular  shortleaf  pine 
beam  of  i8-ft.  span  and  4  inches  broad  to  sustain  a  uniformly 
distributed  load  of  1000  pounds? 

21.  If  a  4-inch  by  4-inch  timber  is  to  carry  a  load  of  50  pounds 
per  foot  what  will  be  the  length  as  a  simple  beam  to  give  the 
maximum  fiber  stress  of  1200  pounds  per  square  inch? 

22.  Design  a  simple  yellow  pine  beam  of  i2-ft.  span  carrying 
concentrated  loads  of  i,  2,  and  3  tons  at  distances  of  3,  6,  and  7 
feet  respectively  from  the  left  support,  and  a  uniform  load  of  f  ton. 

23.  Design  a  rectangular  cast  iron  cantilever  beam  to  carry 
a  load  of  3000  pounds  at  the  end  of  a  4-ft.  span. 

24.  A  rectangular  hemlock   cantilever  beam  8  feet  long  and 
6  inches  deep  is  to  support  a  load  of  200  pounds  at  the  free  end. 
What  should  be  its  width  if  the  weight  of  the  beam  is  neglected? 

25.  A  load  of  500  pounds  is  rolled  over  a  simple  beam  of 
2o-ft.  span.     Find  the  position  of  this  load  for  the  maximum 
bending  moment,  compute  its  value,  and  design  a  longleaf  pine 
beam  to  take  the  load. 

26.  A  round  pin  carries  a  load  of  10,000  pounds  at  the  center. 
It  may  be  considered  as  a  simple  beam  of  6-in.  span.     Find  the 
diameter  of  the  pin  if  the  fiber  stress  is  not  to  exceed  15,000  pounds 
per  square  inch  or  the  shear  to  exceed  7500  pounds  per  square  inch. 

27.  Two  loads  4000  pounds  and  2000  pounds  6  feet  apart  roll 
over  a  simple  beam  12  feet  long.     Find  the  position  of  the  loads 
for  the  maximum  moment  and  determine  its  value.     Design  a 
shortleaf  pine  beam  to  carry  this  load. 

28.  Four  loads,  1000,  2000,  3000,  and  4000  pounds  and  spaced 
2,  3,  and  5  feet  respectively,  roll  over  a  simple  beam  of  i6-ft.  span. 
Determine  position  for  maximum  bending  moment,  and   deter- 
mine its  value.    Design  a  loblolly  pine  beam  to  carry  this  load. 


PROB.]      BEAMS  —  INTERNAL   FLEXURAL   STRESSES      105 

The  following  problems  are  to  be  solved  by  the  use  of 
a  steel  company's  handbook. 

Find  the  missing  terms  by  several  methods  (loads  do 
not  include  weight  of  beam).  /=  16,000  pounds  per 
square  inch. 


G      • 

§ 

x> 

•  -Q 

O  •& 

+* 

. 

o 

o  8  . 

O  Q     • 

• 

OTJ 

**  0^3 

d 

Its 

y 

£~ 

+J    OJ 

i2^ 
S 

Ill 

^J  3  d 

1 

Simple  Beams 


29  

•JQ 

10 
12 

2,OOO 

? 

o 
o 

0 

o 

o 
o 

o 

0 

? 

10"  25# 

21  

14 

O 

28,000 

7 

o 

0 

23 

14 

o 

? 

7 

o 

0 

9"  2I# 

33 

16 

1,000 

10,000 

8 

o 

o 

24. 

16 

? 

20,000 

8 

o 

o 

IS"  9Q# 

?c 

16 

2,000 

? 

8 

o 

o 

15"  8s# 

36 

18 

o 

I  2  ,OOO 

4 

7,000 

9 

•27 

18 

o 

9,000 

3 

9 

7"  20f 

38  

39  
40  

20 

20 
24 

1,000 
1,000 

? 

20,000 

? 

12,000 

6 
8 

o 
o 
o 

0 

o 
o 

12"  3i# 
15"  8o# 

41  

2O 

1,000 

5,000 

6 

6,000 

IO 

? 

42  

2O 

? 

4,000 

8 

5,000 

12 

12"  ioo# 

Cantilever  Beams 


43  .  . 

8 

4,OOO 

o 

o 

o 

o 

? 

44  

IO 

? 

o 

o 

o 

o 

10"  25# 

45  

12 

O 

10,000 

12 

O 

o 

46  

12 

2,OOO 

10,000 

12 

O 

o 

? 

CHAPTER  VII 


w 


STRESSES    IN    SUCH    STRUCTURES    AS    CHIMNEYS, 
DAMS,  WALLS,  AND    PIERS 

76.    KINDS   OF   STRESSES.     For  structures  that  sus- 
tain a  side  thrust  and  a   direct  weight,   as  chimneys, 

dams,  etc.,  there  is  a  combination 
of  direct  and  flexural  stresses. 
The  treatment  given  in  this 
chapter  is  based  upon  the  as- 
sumption that  the  direct  and 
flexural  stresses  act  independ- 
ently of  each  other,  and  that  the 
side  thrust  does  not  cause  ap- 
preciable deflection.  It  is  also 
assumed  that  the  stress  is  pro- 
portional to  the  deformation  and 
that  the  material  of  the  struc- 
ture is  elastic.  For  bearing  on 
soil  this  assumption  may  be  only 
approximately  true. 

When  there  is  no  side  thrust 
and  the  vertical  section  is  sym- 
metrical, the  total  weight  above 
a  horizontal  section  is  resisted 
by  direct  compressive  stresses 
on  the  sections  and  the  unit- 

W 
stress  is/  =  -j  (Fig.  54  (a)).     A 

normal  force  P  produces  flexural 

106 


/A 


Ti  mmm 

(a) 


w 


ART.  77]      ECCENTRIC   LOADS  ON   SHORT   PRISMS          107 


stresses.      A   shearing   stress 

p 
of   s  =  T-T  acts    along  the  sec- 

Ks\. 

tion.  The  bending  moment  due 
to  the  load  increases  the  com- 
pression on  the  opposite  side 
from  the  force  and  decreases  it 
on  the  nearer  side  (Fig.  54  (6)). 
This  lateral  pressure  P  may  be 
due  to  the  wind,  water,  embank- 
ments, etc.  It  may  not  be 
horizontal,  in  which  case  the 
horizontal  and  vertical  compon- 
ents of  the  resultant  of  the 
weights  and  lateral  forces  should 
be  taken  as  producing  the  flex- 
ure! stresses  and  the  direct 
stresses  respectively. 

77.  ECCENTRIC  LOADS  ON 
SHORT  PRISMS.  Let  the  load 
W  have  the  eccentricity  e  (Fig. 
55  (a)).  This  load  may  be  re- 
placed by  its  components,  W, 
Wi,  and  Wz  shown  in  Fig.  55  (6), 
Wi  and  Wz  being  taken  as  acting 
along  the  axis  of  the  prism  and 
equal  in  magnitude  to  W.  The 
two  equal  and  opposite  forces 
W  and  W2  form  a  couple,  the 
moment  of  which  is  We.  The  ef- 
fect of  this  couple  is  to  produce  a 
bending  or  moment  stress.  In  Fig. 
55  (c)  W  and  W2  are  replaced 
by  the  equivalent  moment  C. 


108      STRESSES  IN  CHIMNEYS,  DAMS,  WALLS      [CHAP.  VII 

The  stresses  developed  by  the  eccentric  load  W  will 
be  the  same  as  those  produced  by  its  components  which 
are  the  axial  load  W  and  the  couple  We.  The  stress 
due  to  the  axial  load  is  the  same  at  all  points  of  the  sec- 

W 
tional  area  and   is  -j-  .     The  stress  developed  by  the 

moment  of  the  couple  may  be  considered  as  a  fiber 

j  •  Me       Wee       ,  .     ,      ,. 

stress  and  is  equal  to  -y-  =  —  j—  ,  where  c  is  the  distance 

of  the  most  remote  fiber  from  the  centroidal  axis  of  the 
section,  and  /  is  the  moment  of  inertia  of  the  section 
about  that  axis.  If  r  is  the  radius  of  gyration,  /  =  Ar2. 
Fig.  55  (d)  is  a  free-body  diagram  showing  the  stresses 
developed  by  the  eccentric  load.  The  maximum  stress 
developed  is  in  the  fiber  most  remote  from  the  centroidal 
axis  on  the  side  nearer  the  load  while  the  minimum  stress 
is  developed  in  the  most  remote  fiber  on  the  opposite 
side.  The  maximum  stress  equals  the  sum  of  the 
direct  and  moment  stresses,  and  the  minimum  stress 
equals  the  difference.  Therefore, 

W  .    Wee       W 

f*=A+  —  =  A 

W       Wee       W 


Then,  the  stress  developed  in  the  outside  fiber  is 


78.  ECCENTRICITY  OF  A  LOAD  THAT  WILL  PRODUCE 
ZERO  STRESS  IN  THE  OUTSIDE  FIBER.  If  the  eccentric- 
ity is  increased  /2  becomes  greater  and/i  becomes  smaller. 
After  a  certain  point  is  passed  /i  reverses  if  the  material 
will  take  tension.  If  W  is  a  compression  load,  just 
before  the  tensile  stresses  act,  /i  becomes  zero.  To 


ART.  78]  ECCENTRICITY  109 

obtain  the  eccentricity  e\  that  will  make  j\  zero,  equate 
/i  to  zero.     Then 

W 


where  e\  is  the  greatest  eccentricity  the  load  may  have 
before  tension  is  produced  on  the  side  away  from  the 
compression  load. 

For  a  rectangle  the  eccentricity  to  give  zero  stress  in 
the  outside  fiber  will  now  be  found, 

/        bd*        d2  d 

r   —  ~7  —  — ~n  =  —    and   c  =  - , 
A       12  bd       12  2 

# 

12  _  d 

•**    Cl  =  7  "  6  ' 


Therefore,  as  long  as  a  compression  load  is  kept  on  the 
middle  third  of  a  solid  rectangular  prism  the  stress  over 
the  entire  area  will  be  compression. 
For  a  circle, 


2      64       &        A          d 
r2  =  -j;  =  —    and   c  =  -  > 
Trfir      lo  2 

4 

* 

16  _  d 

ei~~d~s' 

2 

Therefore,  as  long  as  the  load  is  kept  on  the  middle 
quarter  of  a  solid  circular  prism  all  the  stress  will  be 
compression. 


110      STRESSES  IN  CHIMNEYS,  DAMS,  WALLS     [CHAP.  VII 


79-  THE  KERN.  If  the  line  of  action  of  the  load  falls 
within  a  certain  part  of  the  cross  section  of  a  prism, 
all  stresses  in  the  section  will  be  of  one  kind;  and  if  the 
line  of  action  of  the  load 
falls  outside  of  that  area 
the  stresses  will  be  partly 
tension  and  partly  com- 
pression. This  area  is 
called  the  kern  or  kernel. 
In  a  solid  circular  prism 
the  kern  is  a  circular  area 
whose  diameter  is  one- 
fourth  the  diameter  of 
the  prism.  And  for  a 
rectangular  prism  the  kern 
is  a  diamond-shaped  fig- 
ure whose  diagonals  are 
one-third  the  lateral  di- 


f 


FIG.  56. 


mensions  of  the  prism.     (See  Fig.  56.) 

80.  CASE  OF  ECCENTRIC  LOADS  CAUSED  BY  A  COM- 
BINATION OF  THE  WEIGHT  OF  THE  MATERIAL  AND 

LATERAL  PRESSURE.  Call  the  weight  of  the  material 
above  the  section  AB  being  considered,  W,  and  call 
the  lateral  pressure  on  the  prism  above  the  section  P, 
Fig.  57-  The  magnitude  and  direction  of  the  resultant, 
R,  of  these  two  forces  depend  upon  the  forces.  Its 
line  of  action  passes  through  the  intersection  of  W  and  P 
and  intersects  the  section  AB  in  some  point  C  usually 
not  at  the  centroid,  thus  producing  an  eccentric  load  on 
the  section  AB.  The  eccentricity  e  is  DC  and  can  be 
calculated  by  taking  moments  about  the  centroidal  axis 
at  D.  Resolve  R  into  its  vertical  and  horizontal  com- 
ponents, Y  and  H,  at  C  where  it  intersects  the  cross  sec- 
tion for  which  the  stresses  are  to  be  found.  (Fig.  57 


ART.  81]       RESULTANT  OUTSIDE  OF  THE  KERN 


III 


H  produces  a  shearing  unit-stress  along  the  plane  of 

TT 

magnitude  5  =  T-T  ,  and  F  is  an  eccentric  load  producing 
the  stress 

f-ifr.*59 


in  which  the  plus  sign  is  used  to  obtain  the  maximum 
compressive  stress,  /2,  and  the  minus  sign  is  used  to 
obtain  the  minimum  stress,  /i. 

81.  EFFECT  WHEN  THE  LINE  OF  ACTION  OF  THE  RE- 
SULTANT FALLS  OUTSIDE  OF  THE  KERN.  If  the  result- 
ant of  all  loads  above  a  section  of  a  prism  under 
compression  falls  outside  the  kern,  the  minimum  stress 


112      STRESSES  IN  CHIMNEYS,  DAMS,  WALLS      [CHAP.  VII 


1 


/i  =  £  f  I  —  —A  becomes  negative  or  tension  if  the  material 

will  take  tension,  Fig.  58.  For  materials  that  will  not 
take  tension,  such  as  masonry, 
the  joints  on  the  side  opposite  the 
eccentric  load  will  tend  to  open,  but 
failure  will  not  necessarily  follow. 
If  the  structure  is  subject  to  water 
pressure  the  water  may  get  into  the 
cracks  and  produce  an  upward  pres- 
sure which  tends  to  help  overturn 
the  structure.  In  chimneys  and 
walls  where  there  is  no  upward 
pressure  due  to  water  getting  into 
the  cracks  if  they  do  form,  the 
tendency  for  them  to  form  is  not 


FIG.  58. 

so  objectionable.  The  safe 
limit  for  the  compressive 
stress  should  not  be  ex- 
ceeded by  the  maximum 
stress  developed. 

82.  THE  MAXIMUM 
STRESS  WHEN  THE  LINE 
OF  ACTION  OF  THE  RE- 
SULTANT FALLS  OUTSIDE  THE  MIDDLE  THIRD  FOR 
RECTANGULAR  PRISMS  WHICH  TAKE  NO  TENSION. 
The  stresses  will  be  distributed  as  shown  in  Fig.  59  forming 


FIG.  59. 


EXAM.]      STRESSES   IN   CHIMNEYS,   DAMS,  WALLS  113 

a  wedge-shaped  prismatic  stress  volume  BCDEFG,  there 
being  no  stress  on  the  area  ABGH.  The  vertical  com- 
ponent, F,  of  the  resultant  equals  the  summation  of  the 
vertical  resisting  stresses.  The  line  of  action  of  F  passes 
through  the  centroid  of  the  prism  which  is  ^  (CB)  from  CF. 
If  b  is  the  breadth  CF,  the  area  over  which  the  resisting 
stress  is  distributed  is 


If  d  is  the  depth  AC,  and  e  the  eccentricity,  F  is  at  the 
distance e  from  the  edge  CF.     The  length  BC  is 

Id        \ 
3! e].     If  /  is  the  maximum  compressive  stress,  the 

average  on  the  stressed  area  is  -  and  the  total  resisting 


stress  is 


f-KH»-t»«-">' 

(d-2e), 
4Y 


/  = 


EXAMPLES 

i.  Find  the  maximum  stress  at  the  foot  of  a  stone  wall  20  feet 
high  and  4  feet  thick  when  there  is  a  wind  pressure  of  35  pounds 
per  square  foot;  also  when  there  is  a  wind  pressure  of  45  pounds 
per  square  foot  if  the  masonry  weighs  150  pounds  per  cubic  foot. 

Consider  a  portion  of  the  wall  /  feet  long, 

Y=W=  isox  20X4X/  =  i2,oooX/  pounds, 
H=P  =  35 X  2oX/ =  7ooX/ pounds, 

3  X  7°°  X —  =  Ieg  pounds  per  square  inch. 
2X4X  144  XJ 


114      STRESSES  IN  CHIMNEYS,  DAMS,  WALLS      [CHAP.  VII 

The  eccentricity  is  found  by  taking  moments  about  a  centroidal 
axis  of  the  base. 

e  =  70QOX/XIO  =  _7    foot<     Thig  .g  ^  ^  middle  third 

1 2,000  Xl  12 

,  _  F  Fee  _  12,000 X  /  ,  i2,ooo/7X  i2X  24X  12 
~A  I  48X12x2  12  X  12  /x  48X48X48 
=  21  +  1 8  =  3 9  pounds  per  square  inch. 

For  second  part, 

F=i2,ooox/;    H=gooxl;    e  ==QO°Xl°  =-g- 

12,000  12 

This  is  outside  the  middle  third. 

3  (--«)  =  3  (24  ~  9)  =  45  inches. 


.'.     A'  =  45  X  1 2  X  /  =  540 X  /  square  inches. 

,         2  F          2X  12,000  X  /  .      , 

/  =  —j  =  -      — — =  44  pounds  per  square  inch. 

A  540  X  / 

2.  What  should  be  the  thickness  of  a  rectangular  wall  15  feet 
high  to  resist  a  wind  pressure  of  40  pounds  per  square  foot  with- 
out any  tension  in  the  windward  side,  if  the  material  weighs 
140  pounds  per  cubic  foot? 

Let  d  be  the  thickness. 

The  weight  of  each  lineal  foot  isPF=i5XiX  140 d  =  2100 d 
pounds. 

The  wind  pressure  for  each  lineal  foot  is#  =  i5XiX4o  =  600 
pounds, 

600  X  7-5  _  15 


For  zero  stress  e\  =  - . 
6 


2iood         7  d 


d      is     Jo      oo 

1  =     j»  d  =       =  I2-86» 
6      id  7 

d  =  3. 59  feet. 


PROB.]      STRESSES   IN  CHIMNEYS,    DAMS,   WALLS  115 

PROBLEMS 

1.  Find   the   maximum   and   minimum   unit-stress   in   a   rod 
2  inches  in  diameter  under  a  tension  load  of  16,000  pounds  if  it 
is  applied  at  a  point  |  inch  from  the  center  of  the  cross  section. 

2.  What  are  the  dimensions  of  the  kern  in  a  rectangle  3  inches 
by  8   inches?     In  a  hollow  circular  chimney  of  inner  diameter 
8  feet  and  outer  diameter  10  feet? 

3.  In  a  brick  wall  20  feet  high  and  4  feet   thick,  weighing 
115  pounds  per  cubic  foot,  (a)  What  horizontal  wind  pressure 
will  cause  zero  stress  on  the  windward  side  of  the  base  ?     (b)  With 
that  wind  pressure  what  will  be  the  maximum  stress?     (c)  If  the 
wind  pressure  is  40  pounds  per  square  foot  what  will  be  the  max- 
imum  and   minimum   stresses?     (d)  With   a   wind   pressure   of 
40  pounds  per  square  foot  what  will  be  the  stress  on  the  windward 
side  10  feet  above  the  ground? 

4.  A  square  compression  piece  9X9  inches  carries  an  eccen- 
tric load  of  16,200  pounds  so  applied  that  the  stress  on  one  edge 
equals  o.     Determine  the  application  point  of  the  resultant  load. 

5.  What  must  be  the  thickness  of  a  wall  25  feet  high,  weighing 
120  pounds  per  cubic  foot,  if  a  maximum  unit-stress  of  47.2  pounds 
per  square  inch  is  developed  when  the  wind  pressure  is  40  pounds 
per  square  foot?  Ans.   4.5  feet. 

6.  Would  a  brick  wall  30  feet  high,  weighing  120  pounds  per 
cubic  foot,  3  feet  thick  at  the  top  and  4  feet  thick  at  the  base, 
with  one  side  vertical,  be  safe  if  subject  to  a  wind  pressure  of  40 
pounds  per  square  foot? 


CHAPTER  VIII 


GRAPHIC    INTEGRATION* 

83.  DEFINITIONS.  In  Chapter  V  the  relations  be- 
tween the  load,  shear,  and  moment  diagrams  are  given 

as  follows :  the  difference 
between  the  ordinates  at 
any  two  sections  in  the 
moment  diagram  repre- 
sents the  area  in  the 
shear  diagram  between 
the  two  sections,  and 
the  difference  between 
the  two  ordinates  of  the 
shear  diagram  repre- 
sents the  area  in  the  load 
diagram  between  the  or- 
dinates at  the  same  sec- 
x  tions;  thus,  in  Fig.  60, 
LN  in  the  moment  dia- 
gram represents  the  area 
EHIG  of  the  shear  diagram,  and  EF  in  the  shear  dia- 
gram represents  the  area  ABCD  of  the  load  diagram. 
The  first  integrated  curve  is  defined  as  one  in  which  the 
ordinates  represent  the  area  under  a  given  curve.  Thus, 
the  moment  curve  is  the  first  integrated  curve  of  the  shear 
curve,  and  the  shear  curve  is  the  first  integrated  curve 
of  the  load  curve.  Since  the  moment  curve  is  the 

*  For  students  who  have  had  integral  calculus  and  who  do  not  intend 
to  follow  the  graphical  method  of  determining  deflections  of  beams,  this 
chapter  may  be  omitted. 

116 


ART.  84]    OBTAINING  SECOND   INTEGRATED   CURVE     117 

first  integrated  curve  of  the  first  integrated  curve  of 
the  load  curve,  it  is  called  the  second  integrated  curve 
of  the  load  curve.  The  second  integrated  curve  is  one 
in  which  the  ordinates  represent  areas  under  the  first 
integrated  curve.  The  integrated  curve  of  the  second 


FIG.  61.    FIRST  METHOD  OF  GRAPHIC  INTEGRATION 


integrated  curve  is  the  third  integrated  curve.  Simi- 
larly, the  nth  integrated  curve  is  one  in  which  ordinates 
represent  the  area  in  the  (n  —  i)th  integrated  curve. 
The  graphical  method  of  deriving  the  integrated  curves 
from  given  ones  will  be  deduced  before  applying  them  to 
the  theory  of  beams. 

84.  THE  FIRST  METHOD  OF  OBTAINING  THE  SEC- 
OND INTEGRATED  CURVE.  In  the  following  methods,  if 
the  given  areas  are  not  bounded  by  straight  lines,  the 


n8 


GRAPHIC   INTEGRATION 


[CHAP.  VIII 


greater  the  number  of  points  secured  on  the  resulting 
curve,  the  more  nearly  accurate  that  curve  will  be.  The 
equations  representing  the  curves  will  be  deduced  by 
making  the  number  of  points  secured  infinite. 

Let  Fig.  6 1  (a)  be  the  given  curve  of  which  it  is  desired 
to  obtain  the  first  and  second  integrated  curves.     The 


0         A 


FIG.  61.    FIRST  METHOD  or  GRAPHIC  INTEGRATION. 

curve  is  taken  below  the  axis  making  the  area  between 
the  curve  and  the  axis  negative,  which  corresponds  to 
the  load  curves  already  described.  To  obtain  the  first 
integrated  curve,  divide  the  area  into  a  number  of  parts 
as  indicated,  and  measure  each  area  by  any  method. 
Then  from  an  arbitrarily  chosen  axis  O'X'  in  (b)  lay  off 
AB  to  a  selected  scale  to  represent  the  area  AI,  then  lay 
off  CD  to  represent  the  area  A^.  Continue  this  process 
for  the  entire  area  under  curve  (a),  then  connect  by  a 


ART.  84]      OBTAINING  SECOND  INTEGRATED  CURVE      119 

continuous  curve  O'G'D'  all  the  points  thus  obtained. 
This  is  the  first  integrated  curve.  Any  ordinate  to 
this  curve  as  G\G'  represents  the  total  area  accumulated 
from  the  left  end  to  the  section  GH. 

In  physical  problems  there  is  always  a  constant  as  00' 
to  be  added  to  the  area  under  the  curve.  This  constant, 
which  is  called  the  constant  of  integration  is  the  value 
of  the  ordinate  to  the  integrated  curve  at  the  origin. 
In  the  usual  case  it  can  be  determined  by  the  conditions 
of  the  problem.  It  frequently  is  zero.  In  the  shear 
curve,  the  constant  is  the  vertical  shear  at  the  left  sup- 
port as  00'  in  (b).  In  the  moment  curve  it  is  the  moment 
at  the  left  support.  After  determining  the  value  of 
this  constant,  draw  the  axis  OX  which  is  the  true  axis 
of  reference  for  the  integrated  curve  (b) ;  the  true  value 
of  the  function  represented  in  curve  (b)  is  then  H'G'  at 
the  section  GH.  The  area  to  be  considered  in  the 
integrated  curve  is  that  between  the  curve  and  the 
axis  OX. 

To  obtain  the  second  integrated  curve,  divide  the  area 
between  the  curve  O'G'D'  and  the  axis  OX  into  small  parts 
as  indicated  in  Fig.  61  (b),  and  from  some  chosen  axis 
OX  in  (c)  erect,  to  some  scale,  the  ordinate  A'B'  equal  to 
the  area  A\  in  (6),  C'D'  in  (c)  equal  to  the  area  AZ  in 
(b),  and  so  on  until  the  entire  area  in  (b)  is  covered, 
then  the  ordinate  Gi'Hi  represents  the  accumulated  area 
in  (b)  from  the  origin  to  the  section  G'H'.  The  con- 
stant of  integration  will  depend  upon  the  conditions  of  the 
problem.  In  the  illustration  it  is  assumed  to  be  zero. 

Curve  (b)  is  the  first  integrated  curve  of  (a),  and  (c) 
is  the  first  integrated  curve  of  (b)  and  a  second  integrated 
curve  of  (a).  As  long  as  the  constant  can  be  determined, 
a  higher  integrated  curve  can  be  obtained  by  the  fore- 
going method,  the  nth  process  giving  the  nth  integrated 
curve. 


120 


GRAPHIC   INTEGRATION 


[CHAP.  VIII 


85.  THE  SECOND  METHOD  OF  OBTAINING  THE  SEC- 
OND INTEGRATED  CURVE.  The  method  given  in  the 
previous  article  cannot  be  employed  unless  the  constants 
can  be  determined  independently.  For  cases  when  the 


FIG.  62.     SECOND  METHOD  OF  GRAPHIC  INTEGRATION. 


constants  cannot  be  determined  a  second  method  must 
be  combined  with  the  first.  Draw  the  first  integrated 
curve  Fig.  62  (b)  in  the  same  manner  as  in  the  first 
method,  using  the  arbitrary  axis  O'X'.  Project  the 


ART.  85]      OBTAINING  SECOND  INTEGRATED  CURVE      12 1 

areas  AI,  AZj  As,  etc.,  in  (b)  to  the  vertical  axis  X'L  as 
indicated.  If  it  is  desired  to  obtain  the  integrated  curve 
for  (b)  referred  to  any  axis  as  0"X" ,  thereby  assuming 
a  constant  of  integration  0"0',  take  a  pole  Pf  on  that 
axis  a  distance  H  from  the  axis  X'L.  H,  which  is  called 
the  pole  distance,  is  measured  in  the  same  units  and 
to  the  same  scale  as  distances  along  the  axis  OX.  Con- 
nect P'  to  the  ends  of  the  lengths  representing  the  areas 
on  the  line  X'L,  as  P'X',  P'B,  P'C,  etc.  These  lines 
are  called  rays,  and  the  polygon  P'X'L  is  called  the  ray 
or  vector  polygon.  A  i  is  to  be  replaced  by  its  compon- 
ents X'Pf  and  P'B,  A2  by  its  components  BP'  and  P'C, 
etc.  Draw  through  the  mean  ordinate  of  the  area  AI  in 
(a)  a  vertical  line  of  indefinite  length,  as  A\A' .  Also  draw 
through  the  mean  ordinate  of  the  area  Az  in  (a)  the 
vertical  line  A2B',  and  draw  through  the  mean  ordinate 
of  A  3  the  line  A3C'.  Continue  this  process  for  all  the 
elementary  areas  in  curve  (a).  From  the  origin  0' 
in  (c1)  draw  O'A'  parallel  to  P'X'  in  (b),  and  from  the 
point  A '  in  (c'}  where  the  line  O'A '  intersects  the  vertical 
line  through  the  mean  ordinate  of  A\  in  (a)  draw  A'B' 
parallel  to  P'B  in  (b),  and  from  the  intersection,  B'  in 
(c'}  of  A'B'  with  the  vertical  through  the  mean  ordinate 
of  A2,  draw  B'C'  parallel  to  P'C  in  (b).  Continue  this 
process  until  the  entire  polygon  0 'A'B'  —  E'  —  X'  is 
drawn.  This  polygon  is  called  the  string  or  funicular 
polygon  and  the  lines  O'A',  A'B',  B'C',  etc.,  are  called 
strings.  The  ordinate  measured  from  the  horizontal 
axis  O'X"  in  curve  (cr)  represents  the  integrated  area 
of  diagram  (b)  between  the  curve  and  the  axis  OffXff, 
i.e.,  the  ordinate  of  a  second  integrated  curve  of  (a), 
The  constant  0"0'  of  (b)  for  the  axis  0"X"  call  AS. 
Take  any  section  MN,  then  y  in  (cr)  represents  the  accu- 
mulated or  integrated  area  in  (b)  from  the  origin  to  the 
section. 


122  GRAPHIC   INTEGRATION  [CHAP.  VIII 

Proof:  From  the  similar  triangles  P'X'X"  and  O'Aid 

Ai 

in  the  ray  and  funicular   polygons   respectively,  —  = 

ri 

(f^).  from  triangles  P'X'B  and  A'AiBi,  ^  =  ^L?2; 

H  pi 

from  triangles   P'BC   and  B'BiCi,  jj  =  (^lCl)  ;    from 

triangles  P'CD  and  C'CiDi,-=j  =  — — — ;  from  triangles 

ri  P3 

P'DE  and  D'DiEi,  ~  =  *  •  By  clearing  the  above 
equations  of  fractions,  the  following  are  obtained: 

•       Ai'x  =  H  (Aid),  (i) 

(2) 
(3) 
(4) 
Aip*  =  H(DiEi).  (5) 

By  subtracting  the  left  members  of  the  last  four  equations 
from  the  left  member  of  the  first  equation,  and  the 
right  members  of  the  last  four  equations  from  the  right 
member  of  the  first  equation  there  results  the  following 
equation : 

Ai'x  —  Aip i  —  Azpz  —  Aspz  —  A4p4  =  PI  (Aid  —  AiBi 
-  BiCi  -  CiDi  -  DiEi),  and 
Ai'x-  2Ap  =  Hy, 

since    A\d  —  AiBi  —  Bid  —  C\D\  —  DiEi  =  y. 

It  can  be  seen  from  curve  (b)  that  AI'X  —  2  Ap  equals 
the  algebraic  sum  of  the  area  under  the  curve  repre- 
sented by  the  shaded  portion;  therefore,  Hy  equals  the 
area  between  the  axis  and  the  curve  from  the  origin  to 
the  section,  and  by  use  of  a  proper  scale  y  represents 
the  area.  Hence,  (cr)  is  a  first  integrated  curve  of  (b) 
and  is  a  second  integrated  curve  of  (a). 


ART.  85]      OBTAINING  SECOND  INTEGRATED  CURVE     123 

The  greater  the  number  of  parts  into  which  the  area 
under  the  curve  is  divided  the  more  nearly  the  true 
curve  will  the  funicular  polygon  be.  When  the  number 
of  parts  becomes  infinite  the  funicular  polygon  becomes 


FIG.  62.     SECOND  METHOD  or  GRAPHIC  INTEGRATION. 


a  smooth  curve  inscribed  within  the  broken  funicular 
polygon  shown.  This  smooth  curve  is  the  true  inte- 
grated curve  and  can  be  drawn  inscribed  in  the  broken 
funicular  polygon.  In  Fig.  62  (b),  by  choosing  the 


124  GRAPHIC  INTEGRATION  [CHAP.  VIII 

pole  P'  on  the  axis  0"X"  it  assumes  the  constant  of 
integration  equal  to  0"0' ',  and  if  this  is  not  the  true 
value,  curve  (cf)  is  not  the  true  curve,  and  the  process 
so  far  is  only  tentative. 

86.  THE  CONSTANT  OF  INTEGRATION.    The  constant 

of  integration  in  every  case  depends  on  the  conditions 
of  the  given  problem,  and  unless  it  has  been  determined 
from  the  given  conditions  of  the  problem,  the  assumed 
one  as  0"0',  Fig.  62  (b),  generally  is  not  the  true  value, 
but  the  true  value  can  be  determined  by  use  of  the 
curve  (cf).  Whatever  quantity  is  represented  by  the 
area  under  curve  (b)  or  by  its  equivalent,  the  ordinate 
of  the  curve  (cf),  Fig.  63,  there  will  be  two  points  along 
the  X-axis  at  which  the  values  of  the  ordinates  are  known 
or  can  be  determined.  For  beams  these  points  will 
usually  be  at  the  ends,  supports,  or  center.  When  the 
constant  AI  in  Fig.  63  (b)  cannot  be  determined  from 
the  curves  (a)  and  (b)  curve  (c')  must  be  employed. 
For  the  arbitrary  constant  A\  chosen  in  Fig.  63  (b) 
curve  (cf)  shows  a  value  of  zero  at  the  left  end  and  a 
negative  value  X"X'  at  the  right  end.  This  shows  that 
the  negative  area  exceeded  the  positive  area  by  the 
amount  X"X' '.  Suppose  that  the  values  at  each  end 
should  have  been  zero,  the  curve  then  should  end  at  Xn ', 
as  would  the  moment  curve  for  a  simple  beam.  For 
this  to  be  the  case  the  negative  area  in  curve  (b)  must 
be  decreased  and  the  positive  area  increased.  To  ac- 
complish this  the  reference  axis  must  be  lowered,  thus 
making  the  constant  larger  than  the  value  assumed,  A\  . 
The  method  of  obtaining  the  value  of  the  constant 
that  will  make  the  positive  area  equal  to  the  negative 
area  in  (b)  is  to  draw  O'X1  the  closing  line  of  the  funic- 
ular polygon  in  (cf),  then  draw  P'X  in  the  vector  polygon 
parallel  to  O'X1  in  the  funicular  polygon.  Then  through 


ART.  86]        THE  CONSTANT  OF  INTEGRATION 


125 


X  draw  the  horizontal  axis  OX,  giving  XX'  or  00'  in 
(b)  the  true  value  of  the  constant. 


FIG.  63.    To  DETERMINE  THE  CONSTANT  OF  INTEGRATION. 

Proof:  By  lowering  the  reference  axis  in  (b)  to  OX 
the  positive  area  is  increased  the  amount  O"X"XO 
which  equals  /  multiplied  by  XX" .  This  must  equal  H 


126  GRAPHIC  INTEGRATION  [CHAP.  VIII 

multiplied  by  X"X'  in  (<;')  to  have  the  positive  and 
negative  areas  in  (b)  equal.  From  similar  triangles 
P'X"X  in  (b)  and  0'X"X'  in  (cr)  : 

TT  I 


vv"  ~  X'X" 

The  difference  between  the  two  axes  O'X"  and  O'X' 
in  (c')  should  be  added  to  the  ordinates  in  the  original 
curve  to  obtain  the  true  curve.  Proof:  From  similar 
triangles  P'X"X  in  (b)  and  O'SR  in  (c')  : 


°r 

H(RS)  is  the  value  represented  by  the  ordinate  (RS) 
in  (cr)  and  x(XX")  is  the  shaded  area  in  (b)  which  is 
the  increase  over  the  former  value.  Each  ordinate  now 
may  be  increased  to  its  true  value  from  a  horizontal 
axis,  or  the  true  curve  may  be  obtained  by  taking  a 
new  pole  on  the  axis  OX  in  (b)  .  (The  pole  distance  may 
be  changed  if  desired.)  Then  by  using  the  new  pole  P 
and  proceeding  by  the  same  method  used  in  drawing 
(c'),  the  true  integrated  curye  (c)  is  obtained.  The 
student  should  supply  the  proof  that  the  ordinates  in 
(c)  represent  the  area  between  the  curve  and  the  axis 
OX  in  (6). 

87.  UNITS.  The  units  for  the  ordinates  of  the  inte- 
grated curves  will  depend  on  the  units  used  for  x  and  y, 
and  the  units  to  be  used  for  x  and  y  will  depend  on  the 
problem  to  be  solved.  The  unit  for  x  is  the  same  for 
all  curves.  The  unit  for  the  ordinates  of  curve  (b)  is 
the  product  of  the  x  unit  and  the  y  unit  or  the  unit 
formed  by  the  product  xy;  and  that  for  curve  (c)  is  the 
product  of  the  unit  of  the  'ordinate  for  (b)  and  the  x 
unit,  i.e.,  the  unit  formed  by  the  product  x2y.  In 
problems  for  beams  x  will  represent  a  length. 


EXAM.] 


GRAPHIC  INTEGRATION 


127 


EXAMPLES 

i.  By  the  method  of  graphic  integration  draw  the  shear 
and  the  moment  curves  for  a  simple  beam  of  i2-ft.  span  carry- 
ing a  concentrated  load  of  2000  pounds  9  feet  from  the  left 
support. 

The  shear  diagram  is  drawn  in  the  usual  way,  Fig.  64.  Select- 
ing the  pole  P  with  the  pole  distance  equal  to  72  inches  the  moment 


£ 


1000* 


1167* 


2000* 

^_ 


-12- 


(a) 


(b) 


Co) 

FIG.  64. 

curve  (c]  is  drawn,  (p  —  a)  in  (c)  is  parallel  to  (P  —  A}  in  (6), 
(p  -  b)  in  (c)  is  parallel  to  (P  -  B}  in  (6),  and  (£  -  c)  in  (c) 
is  parallel  to  (P  —  C)  in  (6).  To  get  the  moment  at  any  section 
as  at  MN  measure  MN,  using  the  same  ^cale  as  is  used  in  draw- 
ing the  shearing  forces,  and  multiply  by  the  pole  distance,  in 
this  case  72  inches.  This  gives  the  bending  moment  represented 
by  MN  equal  to  66,000  pound-inches. 

2.  Determine  the  vertical  shear  at  the  left  support  (constant 
of  integration)  by  the  graphical  method  for  a  simple  beam  of 
i4-ft.  span  carrying  a  uniform  load  of  500  pounds  per  foot 
and  a  concentrated  load  of  3500  pounds  5  feet  from  the  left 
support. 


128 


GRAPHIC   INTEGRATION 


[CHAP.  VIII 


By  using  the  O'X'-axis  in  Fig.  65  (b)  lay  off  distances  as  AB  to 
represent  the  area  in  the  load  diagram  to  the  left  of  the  section. 
Lay  off  the  concentrated  load  BC  to  the  same  scale  in  pounds. 


Select  any  pole  Pf  with  a  pole  distance  say  60  inches'  and  draw 
(cr)  as  described  in  the  text.  Draw  P'X  in  (b)  parallel  to  the 
closing  line  O'X'  in  (cr).  Then  draw  OX  horizontal  giving  00' 
equal  to  5750  pounds  which  is  the  vertical  shear  at  the  left  support. 


PROBLEMS 

i.   By  the  graphical  methods  draw  the  shear  and  moment 
diagrams  for  the  following  cases: 

(a)  Cantilever  beam  of  Q-ft.  span,  concentrated  load  of  5000 
pounds  at  the  free  end  and  one  of  6000  pounds  at  the  center. 

(b)  Simple  beam  of  i6-ft.  span  with  a  total  uniform  load  of 
18,000  pounds. 


PROB.]  GRAPHIC   INTEGRATION  129 

(c)  Simple  beam  of  i6-ft.  span,  uniform  load  of  1000  pounds 
per  foot  and  a  concentrated  load  of  10,000  pounds  8  feet  from  the 
left  support. 

2.  By  the  graphical  methods  determine  the  vertical  shear  at 
the  left  end  for  each  of  the  following  systems  of  loading: 

(a)  Simple  beam  of  lo-ft.  span  with  a  uniform  load  of  400 
pounds  per  foot  and  a  concentrated  load  of  1000  pounds  3  feet 
from  the  left  support. 

(b)  Simple  beam  of   2o-ft.  span,  concentrated  loads  of  5000 
pounds  7  feet,  2500  pounds  10  feet,  and  10,000  pounds  15  feet 
from  left  support. 

(c)  A  beam  16  feet  long  overhanging  the  right  support  4  feet 
with  a  uniform  load  of  1 500  pounds  per  foot. 

(d)  A  cantilever  beam  of  lo-ft.  span  with  a  uniform  load  of 
300  pounds  per  foot  and  a  concentrated  load  of  800  pounds  at  the 
free  end. 

(NOTE.  The  value  of  the  moment  at  the  free  end  is  zero,  and  that 
at  some  other  point  should  be  calculated  and  laid  off  to  scale  to  draw 
the  closing  line.  The  vertical  scale  for  the  moment  curve  equals  the 
product  of  the  vertical  scale  of  the  shear  curve  multiplied  by  the  pole 
distance.) 


CHAPTER   IX 

DEFLECTION   OF   BEAMS 

ELASTIC  CURVE 

88.  BENDING.  The  elastic  curve  is  the  curve  assumed 
by  the  neutral  surface  of  a  beam  under  load.  The  deflec- 
tion of  beams  can  be  obtained  only  from  the  elastic  curve. 
For  certain  kinds  of  beams  the  reactions,  the  maximum 
shear  and  the  maximum  moment  can  be  obtained  only 
by  the  use  of  the  elastic  curve,  while  for  cantilever, 

simple,     and     overhanging 

o'v  beams    the    reactions,    the 

shear,  and  the  moment  may 
be  obtained  without  its  use, 
also  for  beams  fixed  at  both 
ends  and  loaded  sym- 
metrically the  reactions 
and  shear  may  be  obtained 
without  its  use. 

89.  THE  RADIUS    OF 
CURVATURE    OF    BEAMS. 

In  Fig.  66  let  l\  be  an  ele- 
ment of  length  of  a  beam 
under    load.       The    loads 
FIG.  66.  cause   a   bending   moment 

M  at  the  section  CE.    This 

bending  moment  may  be  considered  constant  over  the 
element  of  length  /i.  The  deformation  due  to  the  shear- 
ing stresses  will  not  be  considered  in  this  discussion. 

130 


ART.  89]      RADIUS  OF  CURVATURE  OF  BEAMS  13! 

The  neutral  surface  AB,  which  was  straight  before  the 
load  was  applied,  is  bent  to  a  curve  under  the  influence 
of  the  bending  moment  M.  Assume  that  a  normal  sec- 
tion of  the  beam  before  bending  remains  a  normal  section 
after  bending,  that  the  moduli  of  elasticity  of  the  material 
in  compression  and  in  tension  are  equal,  and  that  the 
stresses  developed  are  below  the  elastic  limit.  Let  CE 
and  DF  be  two  sections  normal  to  the  beam  and  parallel 
to  each  other  before  bending.  After  bending,  their  planes 
will  intersect  at  some  point  0,  the  center  of  curvature  of 
AB.  Let  the  radius  of  curvature  BO  be  r,  and  the  dis- 
tance of  the  most  remote  fiber  from  the  neutral  surface 
BF  be  c.  Draw  GH  parallel  to  CE,  then  DG  is  the  short- 
ening of  the  top  fiber  in  the  length  l\  and  HF  is  the 
elongation  of  the  bottom  fiber  in  that  length.  For  a 
symmetrical  section  these  deformations  are  equal.  Let 
this  deformation  be  ef  '.  Then  the  unit  deformation  is 

e'  ef 

r  »  and  the  unit-stress  developed  is/  =  TE.     From  the 

/i  k 

f         1      *      Me 
moment  formula  /  =  -y-  • 

e'          Me  e'      Me 

Therefore,  T&=  ~T'    or    T~~^J' 

l\  1  l\      rLl 

Since  h  is  very  small  the  triangles  OAB  and  BHF  may 
be  considered  to  be  similar. 

HF      BF  e'      c 

Hence'  =         >   °r         =     ' 


Mc  _ 

~          (         = 


In  this  equation  M  is  the  bending  moment  for  the 
element  of  length  /i,  r  is  the  radius  of  curvature  of  AB, 
E  is  the  modulus  of  elasticity  of  the  material,  and  /  is 
the  moment  of  inertia  of  the  cross  section  of  the  beam 


132  DEFLECTION  OF  BEAMS  [CHAP.  IX 

about  the  neutral  axis.  It  is  seen  from  the  equation 
that  the  radius  of  curvature  of  a  portion  of  a  beam  varies 
inversely  with  the  bending  moment. 

90.  THE  SLOPE  OF  THE  NEUTRAL  SURFACE.  The 
slope  of  a  curve  at  a  given  point  is  the  measure  of  the 
tangent  of  the  angle  the  curve  makes  with  the  hor- 
izontal axis.  Thus,  in  Fig.  67,  tan  a  is  the  slope  of  the 


FIG.  67. 

curve  AB  at  the  point  P.  In  Fig.  68,  which  is  greatly 
exaggerated,  let  a  be  the  angle  the  neutral  surface  of 
a  beam  at  A  makes  with  the  horizontal,  a'  is  the 
increase  in  the  angle  over  the  element  of  length  AB. 
Since  the  angles  are  very  small  for  ordinary  beams,  the 
tangents  or  slopes  of  the  angles  may  be  considered 
equal  to  the  angles  themselves  without  appreciable 
error;  therefore,  the  increase  in  the  slope  is  equal  to  the 
increase  in  the  angle  measured  in  radians,  and  the  slope 
and  the  angle  may  be  interchanged.  From  the  figure  it 

7  / 

is  seen  that  a'  =  —  ;  hence,  the  increase  in  the  slope  over 
r\ 

the  element  of  length  //  is 

,      l^      M'h'      .  El 

a   =  —  =  -prr-  »   since    r  =  T7  ' 

TI        El  M 

Similarly  the  increase  in  the  slope  a"  over  the  element  of 

M"l  " 
length  BK  is          *    .     The  increase  in  the  slope  over 

ILL 


ART.  90]       SLOPE  OF   THE  NEUTRAL   SURFACE  133 

any  element  of  length  has  a  similar  form.     The  total 
increase  of  the  slope  between  any  two  sections  of  the 


FIG.  68. 


beam  is  obtained  by  adding  all  the  increases  between  the 
sections,  which  is 


KA    I 

~^-  + 


. 


El  El  El 

If  the  slope  at  any  section  is  known,  the  slope  at  another 
section  may  be  found  by  adding  the  increase, 
between  the  two  sections. 


134 


DEFLECTION   OF   BEAMS 


[CHAP.  IX 


91.  THE  SLOPE  CURVE.  The  relation  in  the  last 
article  affords  the  means  of  deriving  the  graphical  method 
of  determining  the  slope  curve  for  a  beam.  The  slope 
curve  (a  curve)  is  one  in  which  the  ordinates  show  the 
values  of  the  slope  at  every  point  along  the  beam.  In 


<af) 


FIG.  69. 

Fig.  69  let  (a')  represent  a  beam,  (c)  the  moment  curve, 
and  (d)  the  slope  curve.  The  element  of  length  /i  is  meas- 
ured along  the  beam  and  u  is  measured  along  the  hori- 
zontal axis.  Since  the  length  /i  is  very  small  it  may  be 
considered  a  straight  line,  then  u  =  l\  cos  a  where  a  is  the 
angle  the  beam  at  the  point  P  makes  with  the  X-axis. 
Since  a  is  almost  zero  cos  a  may  be  taken  equal  to  I, 
and  u  equal  to  /i,  and  then  the  increase  in  the  slope 


between  any  two  sections  becomes 


-^rr 

£Ll 


From  Fig.  69 


(c)  it  is  seen  that  Mu  equals  one  of  the  small  shaded 
portions  of  the  area  under  the  moment  curve,  and  S  Mu 
is  the  sum  of  all  the  small  areas  between  the  sections. 


ART.  93]  THE  ELASTIC  CURVE  135 

Therefore,  to  obtain  the  increase  in  the  slope  of  the 
elastic  curve  of  a  beam  between  two  sections,  divide 
the  area  under  the  moment  curve  between  the  two  sec- 
tions by  the  product  of  the  modulus  of  elasticity  of  the 
material  and  the  moment  of  inertia  of  the  cross  section 
of  the  beam  about  the  neutral  axis. 

If  the  moment  of  inertia  of  the  cross  section  is  not 
the  same  throughout  the  length  of  the  beam,  a  modified 
curve  may  be  obtained  from  the  moment  curve  by  divid- 
ing the  ordinates  in  the  moment  curve  by  El  for  several 
sections,  /  being  the  value  of  the  moment  of  inertia 
at  the  section  where  the  ordinate  to  the  moment  curve 
is  measured,  and  then  taking  the  area  under  this  modi- 
fied curve  for  the  change  in  slope  between  the  two  sec- 
tions. When  the  second  method  of  integration  is  used, 
the  pole  distance  may  be  varied  with  J. 

92.  THE  RATE  OF  CHANGE  OF  THE  SLOPE.    The  rate 

Mu 

of  change  of  the  slope  at  a  section  is  =  —  >  which 

14?  lj  I 

is  the  bending  moment  for  the  section  divided  by  the 
product  of  the  modulus  of  elasticity  of  the  material  and 
the  moment  of  inertia  of  the  cross  section  about  the 
neutral  axis. 

93.  THE   DEFLECTION  OF    BEAMS.    THE    ELASTIC 

CURVE.  In  Fig.  70  (e)  let  APB  represent  the  position 
assumed  by  a  portion  of  the  neutral  surface  of  a  beam 
under  load.  Divide  the  length  of  the  curve  into  the  ele- 
ments //,  /i",  /i'",  etc.  If  at  any  point  P  the  value  of  the 
deflection  y  is  known,  that  for  any  other  point  Q  may  be 
determined  by  calculating  the  increase  in  the  deflection  be- 
tween the  two  points.  Let  the  angles  made  with  the  hori- 
zontal by  the  lengths  //,  //',  /i'",  etc.,  be  a',  a",  a"',  etc. 


136 


DEFLECTION  OF   BEAMS 


[CHAP.  IX 


Then  the  increase  in  y  over  the  length  //  is  y'  =  u'  tan  a! 
where  u'  is  the  horizontal  projection  of  //,  y"  =  u" 
tan  a",  y'"  =  u'"  tan  a",  etc.  The  tangent  of  the  angle 
the  elastic  curve  makes  with  the  horizontal  axis  is  the 


O  r 


FIG.  70 


slope  of  the  curve  at  that  point;  therefore,  the  increase 
in  y  between  the  two  points  is 


=  M'tana'+  u"  tan  a" 


=ua 


u"a" 


+ 


From  Fig.  70  (d)  it  is  seen  that  ua  equals  a  small  shaded 
area  under  the  slope  curve,  and  the  summation  of  all 
the  areas,  2  ua,  equals  the  total  area  under  the  slope 
curve  between  the  two  points.  Therefore,  the  increase 
in  the  deflection  of  the  elastic  curve  of  a  beam  between 
any  two  points  is  equal  to  the  area  under  the  slope  curve 
between  those  two  points.  The  deflection  at  the  sup- 
ports is  usually  known. 


ART.  96]  UNITS   FOR  THE   FIVE   CURVES  137 

94.  THE  RATE  OF  INCREASE  OF  THE  DEFLECTION. 

The  rate  of  increase  of  the  deflection  is  —  =  a.     The 

u 

rate  of  increase  in  the  deflection  at  a  section  is  equal 
to  the  slope  of  the  elastic  curve  at  the  section. 

95.  RELATIONS  BETWEEN  THE  FIVE  CURVES.    Com- 
bining the  relations  between  the  load,  shear,  and  moment 
curves  deduced  in  Chapter  V,  with  those  between  the 
moment,    slope,   and   elastic    curves,   important    results 
are  obtained.     Since  these  five  curves  are  the  principal 
ones  for  beam   problems,   and  since  they  form  a  con- 
tinuous chain  between  the  load  and  the  deflection  of  a 
beam,  they  will  be  referred  to  as  the  five  curves.     The 
relation  existing  between  them  may  be  stated  as  follows : 

Between  any  two  sections  of  the  beam : 

(1)  The  increase  in  the  vertical  shear  equals  the  area 
under  the  load  curve  between  the  sections. 

(2)  The  increase  in  the  bending  moment  equals  the 
area  under  the  shear  curve  between  the  sections. 

(3)  The  increase  in  the  slope  equals  the  area  under 
the  moment  curve  between  the  sections  divided  by  El. 

(4)  The  increase  in  the  deflection  equals  the  area  under 
the  slope  curve  between  the  sections. 

Thus  it  follows  that  the  question  of  determining  the 
elastic  curve  is  one  of  determining  constants  of  integration 
and  of  obtaining  areas  under  curves.  These  principles  will 
be  applied  to  various  kinds  of  beams,  and  the  constants 
determined  and  the  areas  obtained. 

96.  THE  UNITS  FOR  THE  FIVE  CURVES.    In  the  fol- 
lowing discussions    (a)  will   refer   to   the   load   curves, 
(b)  the  vertical  shear  curves,   (c)  the  bending  moment 
curves,   (d)  the  slope  curves,  and   (e)  the  deflection  or 
elastic  curves.     The   modulus  of  elasticity  E  will   be 


138  DEFLECTION  OF  BEAMS  [CHAP.  IX 

taken  in  pounds  per  square  inch,  /  in  (inches)4.  For 
all  the  five  curves  and  for  the  pole  distances  one  inch 
along  the  X-axis  will  represent  m  inches  of  length 
measured  parallel  to  the  beam.  The  scale  of  ordinates 
of  the  curves  will  be: 

Curve  (a)  i  inch  =  w'  pounds  per  inch  run.  I  square 
inch  area  =  w'm  pounds. 

Curve  (b)  i  inch  =  n  square  inches  from  (a)  =  nw'm 
pounds.  I  square  inch  area  =  nw'm2  pound-inches. 

Curve  (c)  i  inch  =  p  square  inches  from  (b)  =  pnw'm2 
pound-inches.  I  square  inch  area  =  pnw'm3  pound- 
(inches)2. 

r-          /j\     •     1.       Q.  square  inches  from  (c)       qpnw'm* 
Curve  (d)  I  inch  =  *    H  pr  -"  =  ^  ^T — 

£Ll  ±L1 

,.  ,    .  ,.  .     ,  pqnw'm* .     , 

which  is  a  ratio.     I  square  inch  area  =        ^r —  inches. 

£Ll 

Curve  (e)  I  inch  =  r  square  inches  from  (d)  =  —        — 

iLl 

inches. 

For  an  illustration  of  the  method  of  determining  the 
scales  of  the  curve  see  Example  I  at  the  end  of  the  next 
chapter. 


EXAMPLE 

i.  What  will  be  the  increase  in  the  slope  from  the  left  end  to 
the  middle  of  a  Q-inch,  2i-pound  I-beam  of  i2-ft.  span  with  the 
concentrated  load  at  the  center  that  will  produce  a  maximum 
fiber  stress  of  16,000  pounds  per  square  inch? 

The  maximum  moment  is  developed  at  the  center  and  is 

M  =  J—  =  16,000  X  18.9  =  302, 400  pound-inches. 

Since  the  moment  increases  directly  from  zero  at  the  end  to  the 
maximum,  the  moment  curve  is  as  drawn  in  Fig.  71.  The  area 


PROB.] 


DEFLECTION   OF   BEAMS 


139 


under  the  moment  curve  to  the  left  of  the  center  is  |  X  72  X 
302,400  =  10,886,400  pound-inches2.  Therefore  the  change  in  the 
slope  over  this  length  is 


10,886,400 
30,000,000  X  84.9 


=  -0043. 


302 


FlG.   71. 


PROBLEMS 

1.  What  is  the  radius  of  curvature  at  the  ends  and  at  the 
center  of  a  3-inch  X  4-inch  stick  8  feet  long  used  as  a  simple  beam 
with  a  load  of  350  pounds  concentrated  at  the  center? 

Ans.    2860  inches. 

2.  What  is  the  change  in  the  slope  from  the  left  end  to  the 
center  in  the  beam  in  Problem  No.  i  ?     What  is  the  change  in  the 
slope  over  the  first  two  feet? 

3.  What  is  the  total  change  in  the  slope  over  the  entire  length 
of  a  cantilever  beam  of  ip-ft.  span  carrying  a  concentrated  load 
of  3000  pounds  at  the  end,  if  the  beam-  is  of  a  standard  I-section 
and  the  stress  does  not  exceed  12,000  pounds  per  square  inch? 

4.  What  is  the  rate  of  increase  of  slope  at  every  two  feet  of 
length  along  the  beams  in  Problems  No.  2  and  3  ? 


CHAPTER  X 

CANTILEVER  AND  SIMPLE  BEAMS  AND  BEAMS  FIXED  AT 
BOTH  ENDS 

97.  CANTILEVER  BEAM,  CONCENTRATED  LOAD  AT 

THE  END.  By  use  of  the  method  analyzed  in  Art.  85 
and  86  the  curves  in  Fig.  72  are  drawn  for  a  cantilever 
beam  with  a  load  W  at  the  end.  From  these  curves  the 
shear,  moment,  slope,  and  deflection  at  any  section  may 
be  scaled  off  directly.  Algebraic  expressions  for  these 
quantities  will  now  be  deduced.  From  the  definition 
of  vertical  shear, 

V  =  -W. 

Since  the  bending  moment  at  the  left  end  is  zero,  that 
at  the  section  AB  is  equal  to  the  area  under  the  shear 
curve  between  the  origin  and  the  section,  which  is  —  Wx. 

M  =  -Wx. 

The  increase  in  the  slope  from  the  left  end  is  equal  to 
the  area  in  the  moment  diagram  from  the  origin  to  the 

Wx2 
section  divided  by  El,  and  equals =rT-    The  free  end 

2  JtltjL 

of  the  beam  deflects.  The  beam  remains  horizontal  at 
the  wall,  thereby  making  the  slope  zero  at  the  wall. 

Wl2 

The  total  area  under  the  moment  curve  is  — This 

2 

divided  by  El  gives  the  total  change  in  the  slope  from 
one  end  to  the  other. 

If  the  slope  at  the  left  end  of  the  beam  is  a\,  it  is 

140 


ART.  97]    CANTILEVER  BEAM,  CONCENTRATED  LOAD   14! 


changed  from  this  value  to  zero  at  the  right  end.     There- 
fore, 


FIG.  72.     CANTILEVER  BEAM,  CONCENTRATED  LOAD 

To  get  the  slope  at  the  section  AB  add  to  ai  the  change 
from  the  left  end  to  the  section  which  gives 

W 


2EI 


142 


CANTILEVER  AND   SIMPLE  BEAMS        [CHAP.  X 


The  change  in  the  deflection  of  the  elastic  curve  is 
equal  to  the  area  under  the  slope  curve.  The  change 
in  the  deflection  is  shown  in  (d)  by  the  area  OO'HI 


-.x  —  area 


which  equals  area  OO'GI  —  area  O'GH  or 

Wl2  2 

O'GH    since    OO'  =  —pj'     Fig.  73  represents  the  area 

O'GH  drawn  to  a  large  scale.     Divide  the  length  x  into 


FIG. 


73- 


a  great  number  of  parts  n  parallel  to  GH  and  let  A  BCD 
be   any   division   such   as   the   pth   from   the   apex   0'. 

/v* 

The  distance  of  this  strip  from  0'  is  p-t  the  breadth  is 

x  W    /px\2  n 

-  »  and  the  depth  is  —  =rT  {  —  )  .     The  area  of  the  strip 

n  2  El  \n  / 

W    p2x* 
then  is  A  BCD  =     „..  ~"|"     In  this  p  represents  any  and 

all  numbers  to  n.     The  area  O'GH,  then,  is  the  sum  of 
all  such  areas  as  A  BCD. 


W 


By  algebra  it  can  be  shown  that 

2^  +  3^  +  n  _. 


*  See  "Higher  Algebra,"  by  John  F.  Downey,  page  373. 


ART.  97]  CANTILEVER  BEAM  143 

W 


_„ 

Area  0  GH= 


2  El 

W 


x*/2n*  +  i>n*  +  n\ 
-;(  -  ^  - 
n*\  6  / 


Since  the  side  O'H  is  a  continuous  curve,  as  n  is  increased 
the  assumed  broken  line  will  approach  the  actual  curve 
and  they  will  coincide  when  n  equals  infinity.  By 
assuming  this  to  be  the  case,  the  actual  area  O'GH  is 

obtained,  and  — and 7— -become  zero;  therefore 

Wx3 
Area  O'GH 


and  Area  00' HI 


6  El 

Wl2x      Wx3 
2  El      6  El 


Since  the  area  under  the  slope  curve  represents  the  change 
in  the  deflection  the  expression  for  the  area  OO'HI  is 
the  increase  in  the  deflection  from  the  left  end  to  the 
section  AB.  In  Fig.  72  (e)  if  the  J^'-axis  were  used  as 
reference,  the  deflection  at  the  left  end  would  be  zero, 
and  the  above  equation  would  give  the  value  of  the  deflec- 
tion at  any  section  AB.  The  axis  of  reference  is  usually 
taken  in  the  position  of  the  neutral  surface  before  any 
load  is  on  the  beam,  for  which  case  the  deflection  at  the 
left  end  is  00'  equal  to  XX',  Fig.  72  (e),  which  equals 
the  total  change  in  the  deflection  over  the  entire  length 
of  the  beam.  This  change  is  obtained  by  letting  x  equal 
/  in  the  expression  for  the  area  under  the  slope  curve, 
WPx  Wy?  WP 


oo>  =  -wp 


= 
y 


3  El 

/2  -y  \A//Y'3  lA/I™ 

v   *v  rvJt  W  v 


2  El      6EI 


144  CANTILEVER  AND   SIMPLE  BEAMS      [CHAP.  X 

which  is  the  equation  of  the  elastic  curve  of  a  cantilever 
beam  with  a  load  W  at  the  free  end.  The  maximum 
deflection  occurs  at  the  free  end  and  is 

WP 

A   =  —  • 

3  El 

98. .'  CANTILEVER  BEAM,  CONCENTRATED  LOAD  AWAY 

FROM  THE  FREE  END.  The  solution  of  the  problem  for 
a  concentrated  load  at  the  end  of  a  cantilever  beam  can 
be  extended  to  cover  the  problem  when  the  load  is 
away  from  the  end.  The  dotted  lines  in  Fig.  72  indicate 
the  extension  of  the  solution  for  the  previous  case  to 
cover  this  case.  The  load,  shear,  and  moment  curves 
would  be  similar  to  those  given.  The  slope  has  the  con- 
stant value  00'  from  the  free  end  to  the  load.  The  addi- 
tional deflection  of  the  free  end  equals  the  area  KJO'O. 
The  student  may  deduce  the  equations  for  this  case. 

99.  CANTILEVER  BEAM,  UNIFORM  LOAD.  In  Fig.  74 
are  '  drawn  the  curves  for  a  cantilever  beam  with  a 
uniform  load.  The  expressions  for  the  values  repre- 
sented by  the  different  curves  at  the  section  AB  the 
distance  x  from  the  free  end  will  be  deduced. 

The  load  per  unit  of  length  of  the  beam  equals  —  w. 

Vertical  shear         V=  —wx. 

Bending  moment  M= 

The  slope  at  the  right  end  is  zero  as  the  elastic  curve 
there  is  horizontal.  The  area  under  the  moment  curve 


to  the  section  AB  is  — —  (see  Art.  97).    The  total  area 
o 

*2¥)J^  *7P)  J^ 

is — — ,  hence  the  total  change  in  the  slope  is  —  -r^-r »  mak- 

D  O  ILL 

>7fjJ3 

ing  the  slope  at  the  left  end  a\  equal  to  ^=r^ ;  therefore, 

O  JLLJ. 

Q1  wl3 

blope  a  ~ 


ART.  99]      CANTILEVER    BEAM,  UNIFORM    LOAD  145 


FIG.  74.    CANTILEVER  BEAM,  UNIFORM  LOAD. 


146 


CANTILEVER  AND   SIMPLE   BEAMS       [CHAP.  X 


The  deflection  at  the  right  end  is  zero.  In  order  to  find 
the  deflection  at  any  section  AB  it  is  necessary  to  obtain 
the  area  under  the  slope  curve  since  the  change  in  the 
deflection  is  equal  to  that  area  which  is  O'HIO.  In 
Fig.  74  W, 

area  O'HIO  =  area  O'GIO  -  area  O'GR 
ivfix 


6  El 


-  area  O'GH. 


FIG.  75- 

Let  Fig.  75  represent  the  area  O'GH  drawn  to  a  larger 
scale.     Divide  it  into  a  great  number  of  strips  n  parallel 

to  GH  as  A  BCD.     The  width  of  each  strip  is  -  •     Let 

n 

this  be  any  strip  as  the  pth  from  0',  then  its  distance 

X  IV     /'tflx^\ 

from  0'  is  p—  and  its  depth  is  'T~pT\~r)'     Therefore 
the  area  A  BCD  equals     l!r  4  •     p  represents  all  numbers 


from  i  to  n.     The  total  area  O'GH  equals  the  sum  of  all 
such  small  areas. 


Area    O'GH  = 
Area    u  uw 


From  algebra  S  (n)s 


w4  +  2  w3  +  w2 


*  See  "  Higher  Algebra,"  by  John  F.  Downey,  page  373. 


ART.  100]     CANTILEVER  BEAM,   VARIOUS  LOADING       147 

Area  O'GH  -     ™*    ("4  +  2  **3  +  O 
'6  EM  4 

WX*   /I    .  __  I 

~  ~~ 


since  —  and  —  5  reduce  to  zero  when  n  equals  infinity. 
2  n          4  n2 

Therefore  area  O'GIO  =  ~  -  -^  •     This  equals  the 

O  £Ll          24  &± 

change  in  the  deflection  from  the  free  end  to  the  section 
AB.  The  total  change  over  the  entire  length  is  obtained 
by  letting  x  equal  /  in  the  expression  for  the  change. 
This  reduces  to 

wl*          wl*          wl* 


wl*  WP 


8  El  ~       8  El 

which  is  the  maximum  deflection  and  occurs  at  the  free 
end.  W  is  the  total  load. 

w/4        wlsx         wx* 
~8^E7  +  6^E7~24£7' 

which  is  the  equation  of  the  elastic  curve  for  a  cantilever 
beam  with  a  uniform  load. 

100.  CANTILEVER  BEAM,  VARIOUS  LOADING.    If  the 

end  projects  beyond  the  uniform  load,  the  load,  shear, 
and  moment  diagrams  will  be  similar'  to  those  of  Fig.  74. 
The  slope  will  be  constant  from  the  free  end  to  the  load 
as  indicated  by  the  dotted  lines  OJKO',  Fig.  74  (d). 
The  additional  deflection  will  be  equal  to  the  area  O'OJK. 
If  the  beam  has  a  concentrated  load  at  the  end  and  a 
uniform  load  the  equations  for  the  two  cases  may  be 
combined.  Any  other  combination  of  uniform  and 
concentrated  loads  may  be  made,  and  corresponding 
equations  derived  similarly  to  the  foregoing  deductions. 


148  CANTILEVER  AND   SIMPLE  BEAMS       [CHAP.  X 

loi.  SIMPLE  BEAM,  CONCENTRATED  LOAD  AT  THE 

CENTER.  The  curves  in  Fig.  76  are  drawn  for  a  simple 
beam  with  a  load  W  at  the  center.  The  vertical  shear 
to  the  left  of  the  load  is 

W 
V  =  -- 

2 

The  bending  moment  to  the  left  of  the  load  is 


M 

M  =  -- 

2 

The  slope  curve  must  pass  through  the  axis  at  the  center 
of  the  beam,  since  the  elastic  curve  is  horizontal  there, 
making  the  slope  at  that  point  equal  to  zero.  The 
change  in  the  slope  from  the  end  to  the  middle  equals  the 
area  to  the  left  of  the  center  in  the  moment  diagram 

Wl2 
divided  by  El,  which  is     ,  ^r;  therefore,  the  slope  at 


Wl2 

the  left  end  is  --  T~^T  »  and  the  slope  at  the  section  AB 
lo  £Li 

to  the  left  of  the  center  is 

Wx2         Wl2 


4  El      i6EI 

The  deflection  at  the  end  is  zero.  The  change  in  the 
deflection  from  the  end  to  the  section  AB  to  the  left 
of  the  center  can  be  obtained  by  calculating  the  area 
under  the  slope  curve  (d). 

Area  OABO'  =  area  OACO'  -  area  O'BC 

/Wl2x        wx*  \          WVx   .    wx* 


\_       WPx 

I  —        ,t  T? r  i 


\i6EI      12  El  I         16EI  '  12  El 

(see  Art.  97).     Since  the  area  is  below  the  axis  it  is 
negative. 

WPx        Wx* 
y       ~i6EI^i2El' 


ART.  101]      SIMPLE  BEAM,  LOAD   AT   CENTER  149 


o  k 


FIG.  76.    SIMPLE  BEAM,  CONCENTRATED  LOAD  AT  THE  CENTER. 


150  CANTILEVER  AND   SIMPLE  BEAMS       [CHAP.  X 

The  maximum  deflection  occurs  at  the  center  and 
equals  the  area  below  the  axis  in  the  slope  diagram, 

which  may  be  found  by  letting  x  equal  -  in  the  expres- 
sion for  the  deflection;  therefore  the  maximum  deflection 
is 

WP 


A  =- 


48  El 


The  foregoing  equations  are  for  a  section  to  the  left 
of  the  center.  Equations  representing  the  curves  to 
the  right  of  the  center  may  be  deduced  similarly  to  the 
above.  The  student  may  derive  those  equations. 

102.  SIMPLE  BEAM,  UNIFORM  LOAD.    The  curves  in 

Fig.  77  are  drawn  for  the  case  of  a  simple  beam  with  a 
uniform  load.     For  the  section  AB  the  vertical  shear  is 


V  =  --  wx. 
2 

The  bending  moment  is 

wlx      wx2 
M=  ----- 

22  x 

The  slope  at  the  center  is  zero;  the  change  in  the  slope 
from  the  end  to  the  center  equals  the  area  under  the 
moment  curve  to  the  left  of  the  center  divided  by  El, 
which  is 

^.-.^      wl2  ^    I  ~vn      ™P      WP      WP 

Area  ODG  =  —  X  -  -  area  OFD  =  —  -  —  =  — 

82  ID        48       24 

(see  Art.  97).  The  slope  is  changed  from  <*i  at  the  left 
end  to  zero  at  the  center;  therefore, 

wl3 


1  24  El  24  El 

The  change  in  the  slope  to  the  section  AB  equals  the 


ART.  102]          SIMPLE   BEAM,  UNIFORM  LOAD  151 


FIG.  77.     SIMPLE  BEAM,  UNIFORM  LOAD, 


CANTILEVER  AND   SIMPLE  BEAMS        [CHAP.  X 


area  OB C  divided  by  El.     By  referring  to  Fig.  78  we  see 
that 

Area     OBC  =  area  ODG  -  area  CBDG 
wl* 


24 


-area  CBDG. 


* 


Area  CBDG  =  area.  CEDG  -  area  BED  =  --      -  x 


-  area  BED  = 


~  area  BED. 


The  ordinate  BE  represents  the  triangular  area  MNQ 
in  Fig.  77  (6). 


A  D777-» 

Area  BED  = 


2X3 


w/l         V 

=  —  [ X] 

6\2  J 


6  \»       4 
Area  OJ5C  =  area  ODG  -  area  CEDG  +  area  BED, 

wl3     wl3 
~  24~T6 
_  wlx2  _  wx3 
~4~      ~& 


wl2x      wl3  _  wl2x      wlx2  _  wx* 
~8~  h  48  "  "8~"    "~4~     '"6" 


The  slope  at  the  section  AB  then  is 
wl3 


a  =  — 


24E7  '  4 £7      6 El 

Since  the  deflection  at  the  end  is  zero,  that  at  the 
section  AB  equals  the  area  under  the  slope  curve  between 


ART.  102]         SIMPLE  BEAM,  UNIFORM  LOAD  153 

the  origin  and  that  section.     In  Fig.  77  (d)  the  deflec- 
tion is  shown  by 
Area  00'BC=  -(area  OO'AC-area  O'AB) 


—  area  0 


'AB\ 


,24  El 
Divide  the  area  O'AB  into  a  great  number  of  strips  n. 

OC 

The  width  of  each  strip  is  -  •      If  the  strip  shown  is 

px 
the  pth  strip  its  distance  from  0'  is  —  >  and  its  depth 

.       wl   (pW\         w     tp*x*\        _u 

is  — ^7^-1  —  ^^7  M-T-j-      The  small   area    then  is 
4  El  \  n2  /      6EI  \nz  I 

wl   /^\         w     /p3x*\   . 

Z^\  w3/  ~  6~EI  I*?/1  m  P  represents  all  num- 

bers to  n,  and  to  obtain  the  total  area  O'BA  all  such 
areas  must  be  added ;  therefore 


4  EM  6  EM 

wlx3    (2  n3  +  3  n2  +  n\ 

r»«v     ~6~     ) 

+  2n3  +  ri 


_ 

2  ^      4  «V 


12  El      24  El 


When   n  is  infinitely  large  —  »  7  —  5  »  and  —  -  reduce  to 

2  n    o  n  A.n2 


zero. 


154  CANTILEVER  AND   SIMPLE   BEAMS       [CHAP.  X 


\24EI      12  El      24  El 

wl3x         wly?         ivx* 


24  El  '   12  £7     24  El 

The  maximum  deflection  occurs  at  the  center  or  when 
x  =  -  in  the  expression  for  y.  Substituting  this  value 
for  x  the  maximum  deflection  is  found  to  be 

~^£/=   ~^El' 
103.  BEAM  FIXED  AT  BOTH  ENDS,  CONCENTRATED 

LOAD  W  AT  THE  CENTER.  A  beam  with  fixed  ends 
has  restraining  moments  at  the  walls  which  keep  the 
beam  horizontal  at  those  points.  These  moments  must 
be  determined  in  the  solution  of  the  problem.  From 
the  symmetry  of  the  beam  the  reactions  at  the  walls 
are  equal,  and  the  restraining  moments  are  equal.  In 
Fig.  79  the  shear  and  moment  curves  are  drawn  in  the 
usual  manner;  then  by  using  the  pole  P'  in  Fig.  79  (c) 
a  slope  curve  (d')  is  drawn.  Since  the  beam  is  hori- 
zontal at  the  ends  and  at  the  center  the  true  slope  curve 
must  pass  through  the  axis  at  those  three  points.  There- 
fore connect  the  ends  by  O'X'.  Now  in  (c)  draw  P'X 
parallel  to  O'X'  in  (df),  and  through  X  draw  the  hori- 
zontal axis  OX  which  is  the  true  moment  axis  giving  the 
moment  OO'  at  the  left  support,  which  equals  XX' ',  the 
moment  at  the  right  support.  Then  with  a  new  pole  P 
on  the  true  axis  in  the  moment  diagram  the  true  slope 
curve  (d)  is  drawn.  To  prove  that  the  slope  in  (d) 
will  be  reduced  to  zero  at  the  right  end,  draw  the  closing 
line  OX  in  (d)  and  draw  OZ  parallel  to  O'X'  in  (d'),  then 
the  angle  ZOX  must  be  equal  to  a  =  a'  in  (d')  because 
Z"X'  and  O'X"  are  both  parallel  to  P'X'  in  (c) ;  i.e.,  hori- 


ART.  103]  BEAM   FIXED   AT   BOTH   ENDS 


FIG.  79.  BEAM  FIXED  AT  BOTH  ENDS,  CONCENTRATED  LOAD  AT  THE 

CENTER, 

zontal.     Y'XO  and  VOX  in  (d)  are  equal  because  Y'X 
and   YO  are  drawn  parallel  to  PX  in  (c). 

X'UXm(c)  =  YOZin(d), 

YOX'm  (d)  =  X'PX'm  (c), 

UX'P'  +  UP'X'  in  (c)  =  X'PX  +  ^rPr^  in  (c) 
=  70^  in  (d)  +  a  in  (rf'), 


156  CANTILEVER  AND   SIMPLE  BEAMS       [CHAP.  X 

/.      YOZ  in  (d)  =  VOX  in  (d)  +  a 

=  YOX  in  (<*)  +  XOZ  in  (d), 
.:     ZOX  in  (d)  =a'm  (df)  =  X'P'X  in  (c). 

To  obtain  the  bending  moment  at  the  walls  and  at  the 
center  of  the  beam,  it  is  known  that  the  slope  is  zero 
at  the  ends  and  at  the  center.  Hence  the  total  change 
in  the  slope  from  the  end  to  the  center  is  zero;  there- 
fore, the  positive  and  negative  areas  to  the  left  of  the 
center  are  equal.  As  the  moment  varies  directly  along 
the  length  of  the  beam  that  at  the  end  is  equal  to 
minus  the  moment  at  the  center.  The  total  change 
in  the  moment  equals  the  area  under  the  shear  curve 

Wl 

to  the  left  of  the  center  which  is  --    Consequently, 

4 

11  •        Wl       ,    , 
the  moment  at  the  wall  is  —  —  and  that  at  the  center 

o 

Wl 

is  -5--     The  deflection  at  the  ends  is  zero.     Since  the 
o 

constants  have  been  determined  the  equations  of  the 
curves  for  a  section  AB  to  the  left  of  the  center  can  be 
written  : 

W 

V=    2  ' 

Wl  .   Wx 


Wlx       Wx* 
~ 


Wlxz  _     Wx* 
I6EI*  I2EI' 

In  order  to  obtain  the  deflection  y  the  area  under  the 
slope  curve  is  determined  by  the  same  method  as  that 
by  which  the  area  under  the  moment  curve  for  a  simple 
beam  uniformly  loaded  was  determined  (Art.  102). 
The  maximum  deflection  is  at  the  center  and  equals 


ART.  104]  POINTS  OF  INFLECTION  157 


the  area  under  the  slope  curve  to  the  left  of  the  center, 
which  is  f 
that  area : 


which  is  found  by  letting  x  =  -  in   the  expression  for 


WP          Wl*  WP 


64  El  '  96.E7          192  El 
104.  POINTS  OF  INFLECTION.    In  Fig.  79  (c)  are 

shown  two  points  marked  /  where  the  moment  is  zero. 
At  these  points  the  moment  changes  from  negative  to 
positive  in  going  toward  the  center  of  the  beam.  The 
stresses  also  change  from  tension  to  compression  in  the 
top  fibers  and  from  compression  to  tension  in  the  bottom 
fibers.  These  points  where  the  fiber  stresses  change  are 
called  the  points  of  inflection  or  points  of  contraflex- 
ure.  Outside  these  points  the  beam  curves  downward, 
and  inside  them  it  curves  upward.  Since  there  are  no 
flexural  stresses  at  the  points  of  inflection,  the  beam 
could  be  hinged  at  those  points  without  affecting  the 
stresses  at  the  other  sections  of  the  beam.*  For  a  beam 
fixed  at  both  ends  with  a  concentrated  load  at  the 
center  the  inflection  points  occur  at  the  two  outside 
quarter  points,  hence  it  may  be  considered  as  a  simple 

beam  of  length  -  with  the  load  W  at  the  center  and  two 

/  W 

cantilevers  each  of  length  -  with  the  load  of  —  at   the 

ends.  The  simple  beam  may  be  considered  as  resting 
on  the  two  cantilevers.  Wherever  the  tensile  stresses 
in  a  beam  are  to  be  taken  by  steel,  as  in  reinforced  con- 
crete beams,  part  of  the  steel  is  bent  down  somewhere  near 
the  inflection  points.  The  inflection  points  are  located 
where  the  greatest  positive  and  negative  slopes  occur. 

*  On  account  of  secondary  stresses  and  horizontal  shear  which  have 
not  yet  been  considered,  the  behavior  of  the  beam  may  be  somewhat 
different  if  hinged  at  the  points  of  inflection. 


158  CANTILEVER  AND   SIMPLE  BEAMS       [CHAP.  X 

105.  BEAM  FIXED  AT  BOTH  ENDS,  UNIFORM  LOAD. 
The  reactions  equal  one-half  the  total  load  for  a  beam 
fixed  at  both  ends  and  carrying  a  uniform  load.  The 
restraining  moments  at  the  ends  keep  the  beam  hori- 
zontal at  these  points.  It  is  also  horizontal  at  the  center. 
In  Fig.  80  curves  (a),  (b),  and  (c)  are  drawn  in  the  usual 
manner.  The  curve  (d')  is  then  drawn  by  using  the  pole 
P'  in  (c).  Connect  O'X'  in  (d'),  then,  in  (c),  P'X  is  drawn 
parallel  to  O'X'  in  (df)  ,  and  OX,  which  is  the  true  moment 
axis,  is  drawn,  giving  the  bending  moments  00'  and  XX' 
at  the  walls.  By  selecting  a  new  pole  P  on  the  moment 
axis  the  true  slope  curve  (d)  is  drawn,  from  which  in 
turn  the  elastic  curve  (e)  is  drawn. 

In  order  to  determine  the  bending  moment  at  the 
wall  and  at  the  center  it  is  known  that  the  slope  is  zero 
at  both  sections,  and,  therefore,  that  the  positive  moment 
area  ABC  equals  the  negative  area  OA  0'  .  The  total 
change  in  the  moment  from  the  end  to  the  center  equals 
the  area  under  the  shear  curve  between  those  two  sec- 

tuft 
tions,  which  is  —  •    By  methods  similar  to  those  already 

o 

given, 

A  ^  A  r\t 

Area  OAO  = 

4  3 

and 

Area  ABC  =  areaABCD  -  areaACD, 


(BC)  =  (AD)  =  W  (area  HIF  in  (b)). 


Area  ABC  =  w 


(DC)  =  -  -  *„ 


Area  0.40'=  area^^C, 

W_Xi_  _  WXi_  m       I ^1  _|__J?ll  _  2X1*  \. 

4  3          6[8  42  J 


ART.  105]  BEAM   FIXED   AT  BOTH   ENDS  159 

Collecting  and  reducing, 

#1=  -  ±  —  Vi2,  which  is  the  distance  from  the  end  to 
the  inflection  point. 


FIG.  80.    BEAM  FIXED  AT  BOTH  ENDS,  UNIFORM  LOAD. 


(AD)  in  (c)  represents  the  area  FHI  in  (6),  and  (O'E) 
in  (c)  represents  the  area  FGO  in  (b) ; 


160  CANTILEVER  AND   SIMPLE   BEAMS       [CHAP.  X 

therefore 

II  V 

(AD)       (DC)2      \z      Xl) 

(Q'E)  ~  (EC}' 
(AD) 


(O'E)  ~  // 


//y 


Hence,  the  bending  moment  (BC)-at  the  center  is  J  of 
—  =  —  '  and  that  at  the  ends  is  ---      The  deflec- 

tion at  the  ends  is  zero. 

Since  all  the  constants  are  determined,  by  methods 
similar  to  those  already  given  the  following  equations 
for  the  curves  at  the  section  AB  are  deduced. 

The  load  per  unit  of  length  of  the  beam  equals  —  w, 

ir       wl 

V  =  --  wx, 

,,         wl2  .   wlx      wx2 

M  =  ---  —  -  » 

12  2  2 

wlzx        wlx2        wx* 
~  + 


wl2x2        wlx3 
~ 


The  deflection  is  the  greatest  at  the  center  and  is  ob- 
tained by  letting  x  =  -  in  the  value  for  y,  or 

=          w/4  Wl3 


384  £/          384^ 

106.  RELATIVE  STRENGTH  AND  STIFFNESS  OF  BEAMS. 
The  strength  of  a  beam  is  proportional  to  the  load  it 


ART.  106]      RELATIVE   STRENGTH   AND   STIFFNESS         161 

will  carry  with  an  assigned  value  of  the  maximum  stress. 
For  a  beam  of  given  section  the  allowable  resisting  shear 
and  resisting  moment  are  fixed  by  the  allowable  stresses. 
If  the  shearing  stress  or  the  deflection  is  not  the  con- 
trolling factor  in  the  design  of  a  beam,  the  strength 

depends  upon  the  allowable  resisting  moment  —  •     The 

0 

allowable  bending  moment  is  equal  to  this  resisting 
moment.  The  strength  of  a  certain  type  of  beam  is 
inversely  proportional  to  the  maximum  bending  moment 
produced  in  the  beam  by  a  given  load.  For  a  beam  of 
length  /,  the  load  W  to  develop  the  fiber  stress  /  may  be 

obtained  by  use  of  the  moment  formula  M  =  —  in  which 

0 

M  is  the  maximum  bending  moment,  here  to  be  expressed 

in  terms  of  W  and  /,  and  -  is  the  section  modulus,  which 

c 

is  fixed  for  a  given  section.  For  example,  in  a  cantilever 
beam  with  a  concentrated  load  at  the  end  M  =  Wl 

(the  sign  being  neglected) ;  hence,  Wl  =  — »  and  W  —  —  ' 

c  cl 

Column  two  in  Table  14  contains  the  expressions  for 
the  maximum  bending  moment  developed  in  the  various 
types  of  beams  given,  and  column  two  in  Table  15  gives 
the  value  of  the  load  to  produce  the  fiber  stress  /.  If 
beams  of  the  same  material  for  the  various  types  given 
are  of  equal  length  and  section,  their  relative  strengths 

will  be  proportional  to  the  coefficients  of  — 7  given  in 

cl 

Table  15,  as  —,  will  be  the  same  for  all  the  beams. 
cl 

The  stiffness  of  a  beam  is  proportional  to  the  load 
necessary  to  produce  a  given  maximum  deflection.  The 
load  W  to  cause  a  maximum  deflection  may  be  obtained 
by  solving  for  W  in  terms  of  the  maximum  deflection  A. 


162 


CANTILEVER   AND   SIMPLE  BEAMS       [CHAP.  X 


For  example,  the  maximum  deflection  of  a  cantilever 
beam   with    a   concentrated   load    at  the  end   is    A  = 

Wl3  El 

(the  sign  being  neglected)  ;   hence,    W  =  3  --  A. 


TABLE   14 

MAXIMUM  MOMENTS  AND   MAXIMUM  DEFLECTIONS 


Kind  of  beam. 

Maximum 
moment 
M. 

Maximum 
deflection 
A. 

I 

1 

f 
'                             1                          i 

—  Wl 

Wl3 

3  El 

W/^in. 

t 

Wl 

Wl3 

L  ik^ 

^ 

2 

SEI 

S                                                                      ' 

Wl 

Wl3 

4 

48  El 

vy/J^in. 

m 

s  Wl3 

r     : 

^ 

8 

384  £/ 

f~ 

\ 

^  8 

192  £/ 

y/, 

W7/        W7/ 

W7/3 

/       / 

y. 

Q       77  7 

f" 

* 

3 

Column  three  in  Table  14  gives  the  expression  for 
the  maximum  deflection  of  each  type  of  beam  there 
shown,  and  column  three  of  Table  15  gives  the  value 
of  the  load  to  produce  the  deflection  A.  If  beams  of 
the  same  material  for  the  types  given  are  of  equal  length 
and  section  their  relative  stiffnesses  will  be  proportional 

7?  T  T?T 

to  the  coefficients  of  -^-  A,  since  -^-  A  is  the  same  for  all 

types,  assuming  equal  deflections. 


ART.  107]     MAXIMUM   STRESS   AND   DEFLECTION  163 


TABLE    15 


LOAD   TO   CAUSE  A   GIVEN   MAXIMUM   STRESS  AND  A   GIVEN 
MAXIMUM    DEFLECTION 


Kind  of  beam. 

Load  TF  to  cause 
stress  /. 

Load  tF  to  cause 
deflection  A 

1 

! 

I 

<•                                  7                               > 

^ 

// 

'•§ 

hi 

3-    -/TA 

'/> 

/•/-, 

« 

\N/l  */  in. 

| 

,      fl 

••'¥* 

IW 

<  ^i 

'/ 
y 

d 

/8 

\ 

1 

4-   - 

48.    f  A 

vy/^in. 

8-f 

76i    f  A 

1 

r 

1 

8    {I 

£/. 

1 
t 

F 
1 

'•   d 

{T 

IQ2.       p   A 
Pr 

g 

J 

W/^/in. 

| 

»-i 

384.   f  A 

f 

Y/ 

General  type. 

«f, 

-^ 

107.  MAXIMUM  STRESS  ANIJ  DEFLECTION.  From 
Art.  1 06  it  is  seen  that  if  a  given  beam  used  as  a  canti- 
lever will  safely  carry  a  given  load  at  the  end,  it  would 
carry  twice  that  load  uniformly  distributed  on  the 
cantilever,  four  times  that  load  if  used  as  a  simple  beam 
with  the  load  concentrated  at  the  center,  eight  times 
that  load  if  used  as  a  simple  beam  with  a  uniform  load, 
eight  times  that  load  if  both  ends  are  fixed  and  the  load  is 
concentrated  at  the  center,  and  twelve  times  that  load  if 
both  ends  are  fixed  and  the  load  is  uniformly  distributed. 
It  is  also  seen  that  if  a  given  load  at  the  end  of  a  canti- 
lever beam  will  cause  a  given  maximum  deflection,  to 


164  CANTILEVER  AND   SIMPLE  BEAMS       [CHAP.  X 

cause  the  same  maximum  deflection  it  will  take  two  and 
two-thirds  times  that  load  uniformly  distributed  over 
the  cantilever,  sixteen  times  that  load  if  the  beam  is 
used  as  a  simple.  beam  with  the  load  concentrated  at  the 
center,  twenty-five  and  one-fifth  times  that  load  if  uni- 
formly distributed  over  a  simple  beam,  sixty-four  times 
that  load  if  concentrated  at  the  center  of  a  fixed-ended 
beam,  and  one  hundred  and  twenty-six  times  that  load  if 
uniformly  distributed  over  a  beam  with  both  ends  fixed. 
In  all  the  above  cases  the  stresses  are  supposed  not  to 
exceed  the  elastic  limit. 

108.  RELATION  BETWEEN  THE  MAXIMUM  STRESS 
AND  THE  MAXIMUM  DEFLECTION.  In  column  2  of 
Table  15  appears  the  maximum  stress  /  developed  under 
the  load  W,  and  in  column  3  the  maximum  deflection. 
By  equating  these  two  expressions  for  the  load  the  rela- 
tion of  the  maximum  deflection  to  the  maximum  stress 
for  a  given  load  W  is  obtained.  Let  a  represent  the  co- 

fT  77  T 

efficients  of^y  >  and  let/3  represent  the  coefficients  of  -=-  A. 
cl  l 

Then  by  equating  the  two  expressions  for  W  there  results 


The  last  equation  gives  the  maximum  deflection  in  terms 
of  the  maximum  stress. 


A=0.63  in. 


(e) 


Scale  Inches 
1 


FIG.  81.    xj*  X  2"  SIMPLE  OAK  BEAM,  SPAN  6',  LOAD  150!  2'  FROM 


LEFT  SUPPORT. 


166 


EXAM.]  CANTILEVER  AND   SIMPLE  BEAMS  167 

EXAMPLE 

Draw  the  elastic  curve  for  a  if -inch  by  2-inch  oak  beam  of 
6-ft.  span  carrying  a  load  of  150  pounds  2  feet  from  the  left 
support. 

Solution:        /  -  ^  -  '*  X  **  2  X  2  =  x  (inch)', 

E  =  1,500,000  pounds  per  square  inch. 

In  Fig.  8 1  the  horizontal  scale  is  i  inch  equals  12  inches.  (On 
the  diagram  the  length  representing  i  inch  is  indicated  at  the 
bottom.)  (a)  represents  the  beam,  (b)  is  the  shear  diagram  in 
which  the  vertical  scale  is  i  inch  equals  100  pounds.  To  draw 
the  moment  curve  the  pole  distance  is  taken  equal  to  24  inches; 
the  vertical  scale  of  the  moment  curve  then  is  i  inch  equals  24  X 
100  =  2400  pound-inches.  In  drawing  the  slope  curve  the  pole 
distance  was  taken  equal  to  20  inches;  the  vertical  scale  of  the 
slope  curve  then  is  i  inch  equals  20  X  2400  -5-  El  =  20  X  2400  -j- 
1,500,000  X  i  =  0.032.  As  yet  the  slope  is  not  known  for  any 
point  along  the  beam;  consequently  an  arbitrary  axis  O'X'  is 
assumed  in  curve  (d)  and  the  pole  P'  taken  with  a  pole  distance 
equal  to  3 1  \  inches  giving  values  of  the  deflection  scale  to  be  i  inch 
equal  to  0.032  X  31!  =  i.o  inch.  With  the  pole  Pt  the  curve 
(e'}  is  drawn.  This  gives  a  deflection  at  the  right  support  equal 
to  0.53  inch.  It  should  be  zero.  The  closing  line  O'X'  in  (ef) 
is  drawn,  and  parallel  to  this  line  the  ray  P'X  in  (d)  is  drawn, 
then  the  true  axis  OX  in  (d)  is  drawn,  and  with  the  pole  P  on  this 
axis,  with  a  pole  distance  of  32!  inches,  draw  the  true  elastic 
curve  (e}.  The  deflections  can  be  measured  directly  from  this 
curve.  The  maximum  deflection  occurs  at  the  point  of  zero 
slope  which  is  31  inches  from  the  left  support.  The  deflection 
at  that  point  is  A  =  0.63  inch. 

If  it  is  desired  to  find  the  maximum  deflection  for  a  5o-pound 
load  divide  the  value  for  150  pounds  by  3;  this  gives  0.21  inch. 
For  any  beam  with  any  concentrated  load  at  the  one-third  point 
this  set  of  curves  can  be  used  simply  by  changing  the  scale  to  agree 
with  the  data  of  the  given  beam. 


i68 


CANTILEVER  AND    SIMPLE   BEAMS        [CHAP.  X 


PROBLEMS 

1.  Draw  the  load,  shear,  moment,  slope,  and  deflection  curves 
and  determine  the  maximum  deflection  and  the  maximum  fiber 
stress  for  the  following  beams: 

(a)  A  4-inch  by  8-inch  timber  beam  used  as  a  cantilever  of 
8-ft.  span  with  a  concentrated  load  of  500  pounds  at  the  end. 

(b)  A   i5-inch,  42-pound,   cantilever   I-beam   of    zo-ft.    span 
carrying  a  uniform  load  of  15,700  pounds. 

(c)  A  simple  timber  beam  of  i4-ft.  span,  8  inches  wide  and 
14  inches  deep  carrying  a  load  of  1550  pounds  concentrated  at 
the  center. 

(d)  Same  as  (c)  with  an  additional  uniform  load  of  200  pounds 
per  foot. 

(e)  A  i2-inch,  31. 5-pound  I-beam  of  i6-ft.  span  when  fixed 
at  both  ends  and  carrying  a  concentrated  load  of  2400  pounds 
at  the  center. 

(f)  An  1 8-inch,  55-pound  I-beam  used  as  a  simple  beam  of 
2o-ft.  span  carrying  a  uniform  load  of  60,000  pounds. 

(g)  A  simple  timber  beam  of  lo-ft.  span,  10  inches  wide  and 
12  inches  deep  carrying  a  uniform  load  of  8000  pounds  and  a 
concentrated  load  of  2000  pounds  at  the  center. 

2.  In  a  test  of  a  i^-inch  by  2-inch  yellow  pine  beam  of  6-ft. 
span  the  following  maximum  deflections   for  the  corresponding 
loads  at  the  center  were  observed: 


Load,  pounds. 

Deflection,  inches. 

50 

.218 

100 

150 

2OO 

•374 

.562 
.718 

250 

•905 

What  is  the  modulus  of  elasticity  of  the  yellow  pine  ? 

Ans.    2,040,000  Ib.  per  sq.  in. 

3.  What  is  the  bending  moment  at  the  walls  and  at  the  center 
of  a  beam  fixed  at  both  ends  of  i6-ft.  span,  and  carrying  a  con- 
centrated load  of  8300  pounds  at  the  center  ?  What  is  the  maxi- 
mum fiber  stress  developed  if  a  zo-inch,  25-pound  I-beam  is 


PROB.]  CANTILEVER  AND   SIMPLE  BEAMS  169 

used?    What  is  the  fiber  stress  4  feet  from  the  walls?    What  is 
the  shearing  stress  at  the  section  4  feet  from  the  walls  ? 

4.  Design  a  longleaf  pine  beam  with  both  ends  fixed  to  carry 
a  uniform  load  of  6000  pounds  on  a  span  of  12  feet.    What  will 
be  the  maximum  fiber  stress  developed  at  the  center  of  the  span  ? 
Locate  the  inflection  points. 

5.  What  steel  I-beam  with  fixed  ends  is  required  for  a  span  of 
20  feet  to  support  a  uniform  load  of  20,000  pounds,  with  a  maximum 
unit-stress  of  15,000  pounds  per  square  inch  ?     Find  also  the  maxi- 
mum deflection. 

6.  For  a  simple  beam  with  a  load  concentrated  at  the  distance 
kl  from  the  left  support,  k  being  a  fraction,  show  that  the  equation 
of  the  elastic  curve  to  the  left  of  the  load  is 


6£/v  6£/v 

7.  Designabeamof  2o-ft.spanto  carry  18,000  pounds,  fixed  ends. 

8.  Calculate  the  maximum  deflection  of  a  steel  bar,  supported  at 
its  ends,  i  in.  sq.,  6  ft.  long,  with  a  load  of  100  pounds  at  its  center. 

9.  A  floor  is  to  support  a  uniform  load  of   100  pounds  per 
square  foot.     The  lo-inch,  2 5 -pound  I-beams  have  a  span  of  20 
feet  and  are  spaced  6  feet  apart  between  centers.     Does  the  maxi- 
mum deflection  of  the  beams  exceed  alo  of  the  span? 

10.  Deduce  the  equation  of  the  elastic  curve  and  the  expression 
for  the  maximum  deflection  for  a  beam  on  which  the  load  varies 
uniformly  from  zero  at  the  ends  to  w  pounds  per  lineal  unit  at 
the  center.     Given   2  (n4  =  [n  (n  +  i)  (6  n3  +  9  n2  +  n  —  i)]  -f-  30. 

Ans.   To  the  left  of  the  center, 


wlx?          wx*          $wPx  .  WP 

y  ~  ' 


24E7      60  Ell      192  El  60  El 

In  the  following  problems  write  the  special  equations 
of  the  elastic  curve  and  obtain  the  maximum  deflections. 

11.  Ai2-in.,3if-poundl-beamusedasacantileverbeamof  20-ft. 
span  and  carrying  a  concentrated  load  of  1000  pounds  at  free  end. 

12.  A  lo-inch,  25-pound  I-beam  used  as  a  cantilever  beam  of 
i5-ft.  span  and  carrying  a  uniform  load  of  500  pounds  per  foot. 

13.  A  20-inch,  65-pound  I-beam  used  as  a  simple  beam  of  24-ft. 
span  carrying  a  concentrated  load  of  20,000  pounds  at  the  center. 

14.  An  i8-inch,  55-pound  I-beam  used  as  a  simple  beam  of 
i5-ft.  span  carrying  a  uniform  load  of  4000  pounds  per  foot. 


CHAPTER  XI 

OVERHANGING,  FIXED  AND  SUPPORTED,  AND 
CONTINUOUS  BEAMS 

109.  OVERHANGING  BEAM,  CONCENTRATED  LOADS. 

In  Fig.  82  are  drawn  the  shear,  moment,  slope,  and  deflec- 
tion diagrams  for  two  concentrated  loads  on  an  overhang- 
ing beam,  W\  at  the  left  end  which  overhangs  the  support, 
and  W2  between  the  supports.  After  drawing  curve  (b), 
curve  (c)  is  obtained  by  use  of  the  pole  P  in  (b).  The 
bending  moment  is  zero  at  the  ends  and  also  at  the  point 
of  inflection  7.  By  use  of  the  pole  P  in  (c)  the  slope 
curve  (d)  is  drawn.  Since  the  value  of  the  slope  is  not 
known  at  any  point,  the  curve  (er)  is  drawn  by  using  the 
pole  P'  in  (d),  thus  assuming  the  slope  at  the  left  end  to 
be  O'A.  The  supports  A'  and  B'  should  be  on  a  hori- 
zontal line.  Therefore,  to  obtain  the  true  elastic  curve 
and  the  correct  value  of  the  slope  at  the  left  end,  connect 
A'B'  in  (er),  then  draw  P'X  in  (d)  parallel  to  A'B'\ 
through  X  draw  the  horizontal  axis  OX,  which  is  the 
true  axis  of  reference  for  the  slope  curve.  This  gives 
the  slope  at  the  left  end  to  be  OA .  Then  by  use  of  any 
pole  P  on  the  axis  OX  in  (d)  the  true  elastic  curve  (e) 
is  drawn.  This  method  is  general  and  may  be  employed 
for  any  system  of  loading  for  cases  in  which  the  beam 
rests  on  two  supports.  If  desired,  the  equations  for 
the  different  parts  of  the  elastic  curve  can  be  obtained 
by  methods  similar  to  those  in  Chapter  X  and  the  ex- 
pressions for  the  maximum  moment,  the  maximum  de- 
flection, and  the  location  of  the  inflection  point  may 
be  obtained. 

170 


FIG.  82.    OVERHANGING  BEAM,  CONCENTRATED  LOADS.          171 


172  BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

no.  OVERHANGING  BEAM,  UNIFORM  LOAD.  In  Fig. 
83  are  drawn  the  curves  for  a  beam  overhanging  both  sup- 
ports and  carrying  a  uniform  load .  The  bending  moment 
is  zero  at  both  ends  and  at  two  points  between  the 
supports.  These  points,  marked  I  in  curve  (c),  are  the 
points  of  inflection.  The  value  of  the  slope  is  not  known 
for  any  point,  so  the  curve  (ef)  is  drawn  by  using  the 
pole  P'  in  (d).  The  supports  should  be  on  the  same 
horizontal  line.  By  connecting  A'  and  B',  the  points  of 
support,  it  is  seen  that  A  'Br  is  horizontal  and  thus  the  true 
axis  in  the  slope  curve  was  assumed  correctly.  The  equa- 
tions of  the  elastic  curve,  and  the  expression  for  the  maxi- 
mum moment,  the  maximum  deflection,  and  the  position 
of  the  points  of  inflection  may  be  determined. 

in.  BEAM  FIXED  AND  SUPPORTED,  CONCENTRATED 

LOAD  AT  THE  CENTER.  For  beams  of  this  kind  the 
values  of  the  reactions  cannot  be  obtained  without  resort 
to  the  elastic  curve.  Referring  to  Fig.  84  the  curves 
marked  by  letters  with  the  subscript  I  are  drawn  as  if 
the  beam  were  a  simple  beam.  To  make  the  beam  hori- 
zontal at  the  right  end  the  restraining  moment  at  that 
end  must  be  great  enough  to  make  the  deflection  /i 
for  the  first  element  of  length  in  Fig.  84  (ei)  equal  to  zero, 
in  which  case  the  slope  <*i  at  the  right  end  is  reduced  to 
zero.  The  resisting  moment  at  the  wall  decreases  the 
left  reaction  and  changes  the  shear,  moment,  slope,  and 
elastic  curves;  from  the  definition  of  bending  moment, 
the  fixing  moment  at  the  wall  is  due  to  a  force  at  the  left 
reaction  equal  to  the  amount  that  reaction  is  decreased 
by  the  fixing  moment.  In  order  to  determine  the  amount 
of  this  force,  a  force  W\  is  assumed  to  be  acting  at  the 
left  end  of  the  beam.  The  shear,  moment,  slope,  and 
elastic  curves  marked  by  the  letters  with  the  subscript  2 
are  drawn  for  the  load  W\.  Since  the  two  ends  remain  on 


174          BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

a  horizontal  line  the  beam  would  curve  upward.  For 
this  case  the  deflection  would  be  /2  for  the  first  element 
of  length  shown  in  the  curve  (e2),  and  the  slope  at  the 
right  end  would  be  a2.  If  the  assumed  force  W\  is  of 


FIG.  84.    BEAM  FIXED  AND  SUPPORTED,  CONCENTRATED  LOAD  AT  THE 

CENTER. 

the  proper  magnitude  so  that  when  the  effect  of  this 
force  is  combined  with  the  effect  of  W  on  the  simply 
supported  beam,  /2  of  curve  (e%)  would  be  equal  to  /i  of 
(ei)  in  order  that  the  deflection  of  the  first  element  of 
length  be  zero.  Or,  expressed  otherwise,  0:2  would  be 
equal  in  magnitude  to  ai  in  order  that  the  slope  at  the 


ART.  in]         BEAM  FIXED  AND   SUPPORTED  175 

right  end  be  zero  under  the  combination.  The  true  force 
W  necessary  to  make  the  moment  at  the  wall  great 
enough  to  bring  the  beam  horizontal  at  that  point  is  to 
the  force  W\  as  the  ratio  of  the  slopes  OL\  to  a2;  therefore 
the  true  force  is 

W'  =  —  Wi. 


By  use  of  the  definitions  and  equations  in  the  previous 
chapters,  the  shear,  moment,  slope,  and  deflection  of  a 
given  beam  are  directly  proportional  to  the  load  causing 
them.  Consequently,  the  reaction  at  the  left  end  of  the 
beam  is  lessened  by  the  amount  W.  The  fixing  moment 
at  the  wall  is  W'l,  or  it  is  equal  to  the  moment  M2 

multiplied  by  the  ratio  —  •     Therefore,  the  fixing  moment 
0:2 


is 


M'=  M2—=Wl  —  l= 

«2  0=2 


For  the  true  reaction  at  the  left  end  of  the  beam  with 
one  end  fixed  and  the  other  supported,  the  left  reaction 
of  the  simple  beam  is  reduced  by  the  amount  W ,  after 
which  the  true  curves  (&),  (c),  (d),  and  (e)  may  be  drawn. 

In  the  foregoing  solution  the  pole  distances  were 
taken  equal,  for  which  case  the  actual  lengths  for  «i  and 
«2  may  be  taken  for  the  reduction  ratio.  If  all  the  pole 
distances  were  not  taken  equal,  the  actual  values  repre- 
sented by  ai  and  0.1  must  be  used  in  the  ratio. 

To  obtain  the  values  of  the  reactions,  the  moment 
under  the  load,  and  the  restraining  moment  at  the  wall, 
it  is  known  that  the  positive  area  and  negative  area  in 
Fig.  84  (J2)  are  equal,  since  the  total  change  in  deflection 
over  the  entire  length  of  the  beam  is  zero.  In  order 
that  these  areas  be  equal; 


176          BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 
«2  =  20:3  (Art.  105). 
«2  -f  0:3  =  jjj  X  -  =          ,(area  in  (c2)  divided  by  El). 


Wl2 

(area  -4-B-X"  in  (ci)  divided  by  £7). 


0:2 


16  16 

'  Mf  =-^-rWl(at  the  wall). 
ID 

M  =  —  Wl  (under  the  load). 

The  inflection  point  occurs  at  the  point  of  zero  moment 
which  is  f\  I  from  the  fixed  end. 

112.  BEAM,  BOTH  ENDS  FIXED,  CONCENTRATED  LOAD 

AT  ANY  POINT.  The  method  of  the  last  article  may  be 
followed  for  a  beam  fixed  at  one  end  and  supported  at  the 
other  with  any  system  of  loading.  When  both  ends  are 
fixed  and  the  loading  is  not  symmetrical  a  method  quite 
similar  to  that  of  the  last  article  may  be  followed,  but 
instead  of  finding  the  fixing  moment  at  one  end  only,  it 
is  necessary  to  determine'  the  fixing  moment  at  both 
ends.  In  Fig.  85  draw  the  curves  (61),  (ci),  (</i),  and 
(ei)  as  for  a  simple  beam.  The  resisting  moments  at  the 


ART.  112] 


BEAM,  BOTH  ENDS   FIXED 


177 


walls  are  proportional  to  ai  and  0.%  if  the  slopes  at  these 
points  are  reduced  to  zero  by  the  resisting  moments. 
Assume  a  force  W\  acting  at  the  left  end,  producing  a 
moment  at  the  right  wall,  and  draw  the  curves  (£2), 
(cz),  and  (dz)  for  this  load.  If  the  assumed  force  were 


0) 


FIG.  85.    BEAM  FIXED  AT  BOTH  ENDS,  CONCENTRATED  LOAD  AT 
ANY  POINT. 

enough  to  make  the  beam  horizontal  at  the  right  end 
a3  would  have  been  equal  to  a2.  If  the  force  were  taken 
at  the  other  end  and  had  been  of  the  right  magnitude 
to  make  the  beam  horizontal  at  the  left  end,  0:3  would 
have  been  equal  to  ai.  Therefore,  the  amount  the  force 
or  reaction  acting  on  the  left  end  is  lessened  to  make 


178  BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

the  beam  horizontal  at  the  right  end  with  the  left  ojie 
simply  supported  is 


and  the  amount  the  right  reaction  would  be  decreased  in 
order  to  have  the  left  end  horizontal  with  the  right  end 
simply  supported  is 


The  sum  of  the  two  reactions  in  every  case  equals  the 
given  load  on  the  beam.  To  have  both  ends  horizontal 
with  the  load  in  the  given  position,  the  right  reaction 
of  the  simple  beam  must  be  decreased  as  the  left  reaction 
is  increased,  by  the  amount 

ray  -  Wt  =  21—2*  WL 


By  making  this  reduction  the  true  shear  diagram  (b) 
may  be  drawn.  The  moment  curve  may  then  be  drawn. 
The  value  of  the  moment  is  not  known  for  any  point, 
but  it  is  known  that  the  slope  is  zero  at  both  ends,  so  by 
use  of  the  pole  P'  in  the  moment  curve,  (d')  is  drawn. 
Through  Pf  in  (c),  parallel  to  O'X'  in  (d1)  P'X  is  drawn 
giving  the  point  X  through  which  the  true  moment 
axis  OX  is  drawn.  By  use  of  the  pole  P  in  (b)  the  true 
slope  curve  (d)  is  drawn  from  which,  by  the  use  of  the 
pole  P,  the  deflection  curve  {e)  is  drawn. 

113.  CONTINUOUS  BEAMS.  The  definitions  and  gen- 
eral equations  given  in  the  foregoing  chapters  are  ap- 
plicable to  continuous  beams  as  well  as  the  general 
method  of  determining  the  elastic  curves.  The  reactions 
of  continuous  beams  are  determined  by  the  use  of  the 
principles  involved  in  determining  the  elastic  curve, 


ART.  113] 


CONTINUOUS  BEAMS 


179 


For  a  given  beam  the  reactions  may  be  determined 
graphically,  but  that  is  left  for  a  more  advanced  treat- 
ment of  the  subject.  In  the  analytical  treatment  of 
continuous  beams  two  spans  are  usually  considered. 
Let  Fig.  86  (a)  represent  two  spans  of  a  continuous 


FIG.  86. 


beam,  and  (6)  and  (c)  represent  the  shear  and  moment 
curves  for  the  same  two  spans.  The  spans  are  taken 
equal  to  /i  and  /2;  the  uniform  loads  are  w\  and  wz  per 
unit  of  length,  with  several  concentrated  loads.  Let  the 
vertical  shear  just  to  the  right  of  the  left  support  be 
Vi  and  that  just  to  the  right  of  the  second  support  be 
V%.  Let  the  bending  moment  at  the  supports  be  MI,  M2, 
and  Ms. 

From  the  definition  of  the  vertical  shear  for  a  section, 


180  BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

the  vertical  shear  for  the  section  AB  in  the  first  span  is 

V  =  Vi  -  wix  -  2  W, 
and  that  for  a  section  in  the  second  span  is 


From  the  definition  of  bending  moment  the  bending 
moment  at  the  section  AB  in  the  first  span  is 

M  =  Mi 


and  that  at  a  section  in  the  second  span  is 


In  the  moment  equation  for  the  first  span  if  x  equals 
M  equals  Mz.     Then 

-  p), 


v        , 

/l  2 


Likewise 


v 

k  2 


By  comparing  the  value  of  Vi  with  the  value  of  the 
vertical  shear  just  at  the  right  of  the  left  support  of  a 
simple  beam  it  is  seen  that  when  there  are  bending 
moments  over  the  supports  the  value  of  the  vertical  shear 

is  increased  by  the  amount  -   —  -,  --  -•      The  value  of 

V\  obtained  shows  that  if  the  bending  moments  at  the 
supports  are  known  the  vertical  shear  to  the  right  of 
each  support  may  be  obtained,  and  the  value  of  the 
vertical  shear  and  the  bending  moment  for  any  section 
then  may  be  found.  If  the  dimensions  of  the  beam  and 
the  load  are  given,  by  use  of  the  shear  and  moment 


ART.  114]      THE  THEOREM   OF  THREE   MOMENTS          181 

formulas  the  stresses  developed  in  the  beam  may  be 
calculated,  or,  if  the  load  is  given,  a  beam  may  be  de- 
signed to  carry  the  load.  The  points  of  inflection  may 
also  be  obtained  by  finding  where  the  bending  moment 
equals  zero.  Consequently,  the  subject  of  the  investi- 
gation and  design  of  continuous  beams  consists  primarily 
of  determining  the  bending  moments  at  the  supports. 

114.  THE  THEOREM  OF  THREE  MOMENTS.    Instead 

of  giving  the  graphical  method  of  determining  the  reac- 
tions of  continuous  beams,  the  theorem  of  three  moments 
will  be  discussed.  In  order  to  determine  the  bending 
moments  at  the  supports  of  a  continuous  beam,  or  the 
relation  between  them,  the  system  of  loading  must  be 
known.  The  relation  between  the  bending  moments  at 
three  consecutive  supports  may  be  deduced  for  various 
systems  of  loading.  (See  Example  7,  Chapter  XII.) 
The  "theorem  of  three  moments"  which  expresses  this 
relation  for  beams  with  uniform  loads  over  each  span  is 


where  Mi,  Mz,  and  Ms  are  the  moments  over  the  sup- 
ports, wi  and  w2  are  the  values  of  the  uniform  loads,  and 


f-        -*-    -f-          -V-         ^ 

FIG.  87. 

/i  and  /2  are  the  spans.  (See  Fig.  87.)  In  applying  the 
theorem  to  a  given  beam,  unless  restrained,  it  is  known 
that  the  bending  moment  at  the  end  support  is  zero. 
As  many  equations  as  there  are  bending  moments  may 


I&2  BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

be  written  and  those  equations  solved  simultaneously 
to  determine  the  bending  moments  at  the  supports. 

For  beams  of  equal  spans  and  a  uniform  load  over 
the  entire  length  of  the  beam  the  theorem  reduces  to 


ILLUSTRATIVE  EXAMPLE 

Given  a  beam  carrying  a  uniform  load  over  three  equal  spans, 
to  determine  the  bending  moment  at  the  supports,  the  vertical 
shear  at  the  supports,  the  reactions,  the  maximum  positive  bend- 
ing moment,  and  the  inflection  points.  See  Fig.  88. 

From  symmetry  Mi  =  Mt  =  o,  and  Mz  =  M3.  Making  these 
substitutions  in  the  theorem  of  three  moments: 


~ 

^ 
10 


T7          wl      wl  _ 

v  i  —  --  H  --  -- 

10  2  IO 


T7      wl   .  wl      6wl 

V  3  =  --  1  --    =   - 
10  2  10 

The  shear  to  the  left  of  the  second,  third,  and  fourth  supports 

is  --  —  j  —  —  >  and   —  -  —  ,  respectively.     The  reaction  equals 
10  2  10 

the  algebraic  difference  of  the  shear  to  the  right  and  to  the  left 
of  the  support.     Therefore, 

K   _  »    _  4  wl 
j\i  —  j\i  —      —  > 
10 


ART.  114]     THE  THEOREM  OF  THREE  MOMENTS         183 

The  maximum  positive  bending  moment  occurs  between  the 
supports  where  the  vertical  shear  is  zero.  To  obtain  the  position 
of  zero  shear  let  V  equal  zero. 


T/ 

V  = 


10 


o. 


±1, 


10 


4^v4/       ^/4A2_ 

—   A  — 
10  10          2  \IO/ 


200 


Since  the  moment  is  zero  at  the  left  support  it  will  be  zero 
again  where  the  shear  areas  above  and  below  the  axis  are  equal, 

o 

which  is  at  the  distance  2x1  or  —  /  from  the  left  support.     This 

10 

gives  the  inflection  point.     In   the  middle  span   the  shear  is 


FIG.  88. 


zero  at  the  center.     The  area  under  the  shear  curve  to  the  center 


;   then  the  bending   moment    at  the   center   is 


is    --  X  - 
22 

---  1  --  =  -  .     The  inflection  points  occur  where  the  bend- 
10        4          20 

ing  moment  passes  through  zero.    Therefore,  the  distance  from 


184          BEAMS,  OVERHANGING,  FIXED,  ETC.      [€HAP.  XI 
the  second  support  to  the  inflection  point  is  found  from: 

W?     ,    Wl  WXi 


The  slope  and  the  deflection  may  be  obtained  by  graphical  methods 
or  by  the  calculation  of  areas  as  is  done  in  the  previous  chapters. 

In  Fig.  88  (6)  is  drawn  the  shear  diagram  and  in  Fig.  88  (c)  is 
drawn  the  moment  diagram  for  this  beam. 

By  a  method  similar  to  that  employed  in  the  fore- 
going example  the  coefficients  for  wl  for  the  vertical 
shears  at  each  side  of  the  supports  for  continuous  beams 

TABLE  16 

COEFFICIENTS  OF  wl  FOR  THE  VERTICAL   SHEAR  AT  THE 
SUPPORTS  OF  CONTINUOUS  BEAMS. 


1 

-fif 

_^_ 

8 

Spa 

t_ 

1                           it' 

3"" 

n 

•ft       -; 

foilo" 

TojTo" 

n 

1           -, 

*$<           1 

A 

~1  ^   " 

ft 

r         i 

17;  15 

28!  28 

_13|M 
28!28 

- 

i         ?l 
.JfiUI 

28128 

28 

£z 

1            str 

1           •$< 

3 

—  4 

* 

23|20             __J 
38|  38                   : 

J9|18_ 
38|38 

__20|23_ 
38[38 

38~" 

t  —  j— 

~±  7  

-±  r— 

~T~ 

tr- 

"*                 J 

p        *• 

4L                      63  1  55 
104              ~104!lOl 

>|<           1 
49!  51 
104;  104 

n<         i 

53  !  53 
104J104 

*P 

J1IJL        _ 
1011104 

J5!J3_ 
1041104 

_  JL 

r« 

t            1            it* 

1             >1V 

—  1  4«r  

—  4<  1  — 

i       >T 

carrying  uniform  loads  over  the  entire  length  were  ob- 
tained as  given  in  Table  16.  The  negative  coefficients 
of  wl2  for  the  bending  moments  at  the  supports  were  also 
obtained  as  given  in  Table  17. 

Tables  16  and  17  may  be  extended  in  the  following 
manner:  By  following  down  to  the  right  or  to  the  left  a 
line  of  similar  supports  for  the  different  spans,  to  obtain 


ART.  115]  HINGING  POINTS  FOR  CONTINUOUS  BEAMS    185 

the  coefficients  for  a  beam  having  an  odd  number  of 
spans,  as  five,  for  the  second  support,  the  moment  coeffi- 
cient is  ^V  The  4  is  obtained  by  adding  the  3  of  ^  to 
the  i  of  TV-  The  38  is  obtained  by  adding  28  of  ft-  to 
the  10  of  TV.  This  method  may  be  employed  for  any 

TABLE  17 

COEFFICIENTS  OF    -rf  FOR  THE  BENDING   MOMENT  AT 

THE  SUPPORTS  OF  CONTINUOUS  BEAMS.  No. 

Spans 

0  K  0 


-1 H< 1 H 

Ho Ho o 

a j ° o 


^28 ^8 ^28 0 

o 2 2 J 4 


^  3/^8  Hj 


Moi M(M 1Mo4 Q 

°  G 


^-.—^^-^^-t-^^-^-^ 


support  of  any  beam  with  an  odd  number  of  spans. 
For  a  beam  with  an  even  number  of  spans,  as  four,  the 
coefficient  is  gV  The  3  is  obtained  by  multiplying  the 
i  of  TV  by  2  and  adding  the  i  of  \.  The  28  is  obtained 
by  multiplying  the  10  of  TTo  by  2  and  adding  the  8  of  \. 
This  method  can  be  followed  for  extending  either 
Table  16  or  17. 

115.  HINGING  POINTS  FOR  CONTINUOUS  BEAMS.    If 

a  continuous  beam  is  to  be  made  of  several  parts,  it  is 
necessary  to  know  at  what  points  the  various  parts 
should  be  hinged,  in  order  that  the  "  continuous  "  effect 
may  be  secured,  as  a  continuous  beam  is  stronger  than 
several  simple  beams  over  the  various  spans.  Any 
given  continuous  beam  may  be  hinged  at  the  inflection 
points,  and  the  bending  moment  would  be  unchanged 
along  the  beam. 


i86       BEAMS,  OVERHANGING,  FIXED,  ETC.    [CHAP,  xi 

An  economical  method  is  to  hinge  the  beam  at  such 
points  as  to  make  the  maximum  negative  bending 
moments  at  the  supports  and  the  maximum  positive 
bending  moments  in  the  spans  equal  in  magnitude.  The 
portion  of  the  beam  between  the  hinges  in  a  span  acts 
as  a  simple  beam  and  the  portions  from  the  support  to 
the  hinges  act  as  cantilever  beams.  For  the  case  of 


Hinge-. 


Wf/m. 


FIG.  89. 

uniform  loads  and  equal  spans,  Fig.  89,  each  hinge 
carries  one-half  the  load  on  the  intermediate  length. 
If  /  is  the  length  of  one  span,  /i  the  distance  between  the 
hinges,  12  the  distance  from  the  support  to  a  hinge,  and 
w  the  load  per  unit  of  length,  the  maximum  bending 

7   2 

moment  in  the  center  is  -~-  and  the  maximum  bending 

o 

moment  at  "the  support  is  —  (  —  -  -\  --  —  1-  For  equal 
maximum  bending  moments 


II 

---  T=  =  .14644  /. 
2       -v/8 


-    -J=  =  —r=  .70712  . 

4  +  v  8        v  2 


PROB.]         BEAMS,   OVERHANGING,   FIXED,   ETC.  187 

From  this  relation  the  maximum  bending  moment  is 

found  to  be 

,,    _  wlS  _  Wl 
~8~   =  76" 

Thus  it  is  seen  that  if  the  beams  are  hinged  at  the  proper 
points  the  efficiency  is  increased  from  thirty  to  sixty  per 
cent.  To  use  beams  hinged  in  this  way  they  should  be 
fixed  at  the  end  supports.  With  uniform  load  the  beams 
would  remain  horizontal  at  the  supports,  but  if  the  load 
is  not  uniform  at  any  time,  the  beam  should  be  fixed 
at  all  the  supports.  Two  lengths  of  beams  could  be 
used,  one  length  about  three- tenths  the  length  of  one 
span,  to  be  used  over  the  supports,  and  the  other  length 
about  seven-tenths  the  length  of  one  span,  to  be  used 
between  the  hinges.  If  the  loads  are  concentrated  at 
the  middle  of  the  spans  the  lengths  should  be  made 
equal. 


PROBLEMS 

1.  Draw  the  shear,  moment,  slope,  and  elastic  curves  for  a 
9-inch,    2i-pound  I-beam  of  length    20  feet,  overhanging  each 
support  4  feet,  carrying  concentrated  loads  of  10,000  pounds  at 
the  left  end,   12,000  pounds  8  feet  from  the  left  support,  and 
15,000  pounds  at  the  right  end.     From  the  curves  determine  the 
deflection  at  each  load  and  the  maximum  deflection. 

2.  What  are  the  maximum  shearing  and  fiber  stresses  developed 
in  the  beam  of  Problem  No.  i  ? 

3.  Design  a  rectangular  Washington  fir  beam  18  feet  long, 
overhanging  one  support  4  feet,  to  carry  a  total  uniform  load  of 
9000  pounds.    The  shearing  unit-stress  is  not  to  exceed  100  pounds 


l83  BEAMS,  OVERHANGING,  FIXED,  ETC.      [CHAP.  XI 

per  square  inch,  the  maximum  fiber  stress  is  not  to  exceed  1200 
pounds  per  square  inch,  and  the  maximum  deflection  is  not  to 
exceed  sl&  of  the  span  between  the  supports. 

4.  Draw  the  shear,  moment,  slope,  and  elastic  curves  of  a 
beam  fixed  at  one  end  and  supported  at  the  other,  of  length  /, 
carrying  a  uniform  load  of  w  pounds  per  lineal  inch,  and  determine 
the  value  of  the  reactions,  the  restraining  moment  at  the  wall, 
the  maximum  positive  moment,  and  the  elastic  curve. 


WX*          wlx 

~r 


24  El       i6EI       48  El 

5.  Draw  the  shear,  moment,  slope,  and  elastic  curves  for  an 
8-inch  by  lo-inch  beam  of  12 -ft.  span  fixed  at  one  end  and  sup- 
ported at  the  other,  carrying  a  concentrated  load  of  8000  pounds 
7  feet  from  the  restrained  end.     What  are  the  maximum  shearing 
and  fiber  stresses  developed  in  the  beam? 

6.  Solve  Problem  No.  5  if  both  ends  are  fixed. 

7.  A  continuous  beam  of  two  spans  carries  a  load  of  100  pounds 
per  foot  over  one  span  of  1 2  feet  and  200  pounds  per  foot  over  the 
other  span  of  8  feet.     Determine  the  moment  at  the  middle 
support  and  the  reactions. 

Ans.  Mz  —  20,640 lb.-in.,  Ri  =457  lb.,  R%  =  1758  lb.,  R3  =  585  lb. 

8.  Determine  the  bending  moment   at  the   middle   support 
and  the  maximum  positive  bending  moments  in  each  span  of  a 
beam  24  feet  long,  one  span  being  10  feet  and  the  uniform  load 
for  that  span  24,000  pounds,  the  other  span  being  14  feet  and  the 
uniform  load  for  that  span  28,000  pounds.     Select  the  proper 
I-beam  for  this  loading. 

9.  Select  the  proper  continuous  I-beam  to  carry  a  uniform  load 
of  144,000  pounds  uniformly  distributed  over  six  spans  of  12  feet 
each. 

10.  If  the  beam  of  Problem  No.  9  were  fixed  at  the  supports 
and  hinged  so  as  to  make  the  bending  moment  at  the  supports 
equal  to  that  at  the  middle  of  the  span,  what  I-beam  would  be 
required  ?    What  would  be  the  length  of  each  section  ? 


CHAPTER  XII 

ELASTIC   CURVE   OF  BEAMS  DETERMINED  BY  THE 
ALGEBRAIC   METHOD* 

116.  THE  ALGEBRAIC  RELATIONS  BETWEEN  THE  FIVE 
CURVES.  As  deduced  in  Art.  89  the  expression  for  the 
radius  of  curvature  of  a  beam  is 

El 


where  E  is  the  modulus  of  elasticity,  /  is  the  moment 
of  inertia  of  the  cross  section  about  the  neutral  axis, 
and  M  is  the  bending  moment.  The  algebraic  expression 
for  the  radius  of  curvature  for  a  curve  as  deduced  in 
the  calculus  is 


r  = 


(2) 


where  x  and  y  are  the  coordinates  of  the  point  of  the 
given  curve,  for  which  r  is  the  radius  of  curvature  and 

•f  is  the  slope  of  the  tangent  to  the  curve  at  the  given 

point.     For   beams   the   X-axis   is   horizontal   and    the 
F-axis  is   vertical,   and   since   the  slope  of   the   elastic 

curve  is  small  at  all  points  of  the  beam  the  value  of  f  -^- 

*  This  chapter  introduces  the  calculus  method  for  the  only  time  and 
is  intended  only  for  students  who  have  had  courses  in  differential  and 
integral  calculus. 

180 


100     ELASTIC  CURVE  —  ALGEBRAIC  METHOD     [CHAP.  XH 

generally  is  very  small,  and  in  comparison  with  i  may  be 
neglected,  with  which  approximation 


dr* 

By  substituting  this  value  of  r  in  equation  (i)  there 
results 

£/g  =  Jf.  (4) 

By  combining  this  relation  with  those  given  in  Chapters 
V  and  IX  the  following  values  for  the  section  distant  x 
from  the  origin  result: 

The  ordinate  to  the  elastic  curve  is 

*=/(*)-  (a) 

The  slope  of  the  elastic  curve  is 

<•=£.  o.) 

The  bending  moment  is 


(C) 
The  vertical  shear  is 


The  load  per  unit  of  length  is 


= 
dx  ~  dr*  dx*  dx* 

If  the  value  of  any  one  of  the  variables  is  known  for 
the  above  equations,  the  values  of  those  lower  in  the 
scale  may  be  determined  by  differentiation  as  indicated. 
but  usually  it  is  necessary  to  start  with  the  lower  equa- 
tions and  derive  the  higher  ones  by  integration.  In 
problems  concerning  beams  the  operation  of  integration 
between  definite  limits  is  not  generally  applied,  con- 
each  operation  introduces  a  constant  of 


Aft.  ittSl     ALGEBRAIC  RELATIONS  OF  FIVE  CURVES     191 

integration  which  must  be  determined  born  the  known 
conditions  governing:  the  case-  For  deriving  higher 
curves  the  equations  may  be  written  in  the  following 
form: 

The  load  per  unit  of  length  is 

=  (i) 

The  vertical  shear  is 


The  bending  moment  is 

M  =  fvd*  +  Ml=ffwd**+fvjx  +  Mi.       (3) 

The  slope  is 
_, 


The  deflection  is 


5 

The  method  of  evaluating  these  expressions  will  be 
given  later. 

The  latter  set  of  equations  is  the  one  to  be  employed 
in  determining  the  elastic  deflections.  Any  one  of  the 
equations  may  be  used  to  start  with,  if  the  variables 
can  be  expressed  in  terms  of -jr.  The  load,  shear,  and 
moment  equations  can  usually  be  written  by  applying 
the  definitions.  If  the  moment  equation  is  used  to 
start  with,  one  integration  and  the  determination  of 


IQ2     ELASTIC  CURVE  —  ALGEBRAIC  METHOD     [CHAP.  XII 

constant  of  integration  are  avoided,  but  since  these 
operations  are  of  the  simplest  in  calculus  there  is  no 
advantage  in  starting  with  any  other  than  the  load  or 
shear  equation.  The  constants  should  be  determined 
as  they  appear  if  convenient.  In  the  case  of  concen- 
trated loads  the  equation  of  the  load  curve  is  zero,  and 
the  shear  curve  probably  would  be  the  best  with  which 
to  start. 

117.  THE  CHOICE  OF  COORDINATE  AXES.    In  the 
deduction  of  the  formula  for  the  radius  of  curvature 

Tf-r 

r  =  -jj »  the  X-axis  was  taken  parallel  to  the  axis  of 

the  beam  before  bending,  and  the  F-axis  at  right  angles 
to  the  X-axis.  The  origin  may  be  chosen  arbitrarily, 
and  for  some  particular  cases  it  is  more  convenient  to 
take  the  origin  at  the  center  of  the  beam,  but  in  this 
book  the  X-axis  will  be  taken  to  coincide  with  the  axis 
of  the  beam  before  the  beam  is  bent,  and  the  F-axis 
will  be  taken  at  right  angles  to  the  X-axis  at  the  left 
end  of  the  span  under  consideration.  In  the  solutions 
the  proper  algebraic  signs  should  be  observed. 

118.  THE  CONSTANTS  OF  INTEGRATION.    In  all  cases 

an  approximate  diagram  of  the  deflected  beam  will 
be  of  value  in  determining  the  constants  of  integration. 
For  problems  in  the  determination  of  the  deflection  of 
beams,  the  constant  of  integration  for  any  curve  is  the 
value  of  the  variable  at  the  origin,  as  here  treated. 
Thus,  Fi,  introduced  in  equation  (2),  Art.  116,  is  the 
value  of  the  vertical  shear  at  the  origin.  See  Fig.  90. 
Mi  introduced  in  equation  (3)  is  the  value  of  the  bending 
moment  at  the  origin;  ai  introduced  in  equation  (4) 
is  the  value  of  the  slope  at  the  origin;  y\  introduced  in 
equation  (5)  is  the  value  of  the  deflection  at  the  origin. 


ART.  119]  CONSTANTS  OF  INTEGRATION 


193 


119.  DETERMINATION  OF  THE  CONSTANTS  OF  INTE- 
GRATION. If  the  values  of  the  constants  can  be  deter- 
mined, they  may  be  inserted  into  the  equations  at  once; 
thus,  Vi  and  MI  can  be  determined  in  many  cases  at 
first.  For  other  cases  it  may  be  known  where  the 
shear  is  zero,  and  then  the  value  of  zero,  for  V  and  the 
corresponding  value  of  x  may  be  substituted  in  equation 
(2)  to  give  the  value  of  V\.  If  the  position  of  zero 
bending  moment  is  known,  the  value  of  M  and  the 
corresponding  value  of  x  substituted  in  equation  (3) 
will  give  MI.  Likewise  to  determine  a\  it  may  be 
known  where  the  slope  is  zero,  i.e.  where  the  beam  is 
horizontal,  and  that  value  of  a  and  the  corresponding 
value  of  x  substituted  in  equation  (4)  will  give  the 


194     ELASTIC  CURVE  —  ALGEBRAIC  METHOD     [CHAP.  XII 


value  of  a\.  For  determining  y\  it  is  known  where 
the  deflection  is  zero,  which  is  at  the  supports  for  the 
usual  cases. 

ILLUSTRATIVE  EXAMPLE 

Deduce  the  equation  of  the  elastic  curve  and  the  value  of  the 
maximum  deflection  for  a  cantilever  beam  with  a  concentrated 
load  at  the  end.  See  Fig.  91. 


FIG.  91. 

The  load  per  unit  of  length  =  o. 
V  =  Vl  =  -W, 

M  =  _  J  Wdx  -f  Ml  =  -  Wx  +  Mi, 

M  =  o  when  x  =  o,    .*.  MI  =  o.     (Zero  moment), 

«  =  -  1=7  f  Wxdx  +  <*!=-  -±-Wx*  +  ai. 
£Ll  J  2  rLl 

The  slope  a  equals  zero  when  x  equals  /,  as  the  beam  is  horizontal 
at  the  wall,  therefore 


Wl2 
2  El 


Wl2 
2  El 


(Zero  slope) , 


Wfx 


y  =  o,  for  x  =  /, 
WP       WP 
6EI  +  2EI 
Wx*   .  Wl2x 


o  =  — 


w/3 

-  -^~-    (Zero  deflection), 
3  El 


wi* 


6  El      2  El 
The  maximum  deflection  occurs  where  x  =  o,  and  is 

WP 
A  =  — 


ART.  1 19]  CONSTANTS  Of  INTEGRATION 


When  there  are  concentrated  loads  on  the  beam  the 
shear  equation  changes  at  every  concentrated  load, 
consequently  the  equations  for  the  moment,  slope,  and 
elastic  curves  have  different  expressions  on  each  side 
of  the  load,  and  for  each  of  the  curves  there  is  one  more 
equation  than  there  are  concentrated  loads  on  the 
beam.  Care  should  be  taken  in  substituting  values  of 
x  in  these  equations  to  see  that  the  equation  is  true  for 
the  particular  value  of  x  used.  With  concentrated 
loads  the  two  sections  of  the  beam  on  each  side  of  the 
load  have  a  common  slope  at  the  load,  and  also  a  com- 
mon deflection.  For  continuous  beams  and  overhanging 
beams  the  two  sections  on  each  side  of  a  support  have  a 
common  slope  and  a  common  deflection  at  the  support. 


FIG.  92. 


Thus,  in  Fig.  92  the  portions  of  the  elastic  curve  AB  and 
BC  have  a  common  tangent  (i.e.  a  common  slope)  and 
a  common  deflection  at  the  point  B.  Also  the  portions 
BC  and  CD  have  a  common  tangent  and  a  common 
deflection  at  the  point  C.  The  beam  is  fixed  at  the 
point  D,  hence  the  slope  of  the  portion  CD  is  zero  at 
the  wall. 

The  following  principles,  then,  may  be  used  in  the 
determination  of  the  constants  of  integration: 

(a)  The  section  of  zero  vertical  shear  can  be  obtained 
by  drawing  the  shear  diagram,  and  if  it  occurs  at  a 


196     ELASTIC  CURVE  — ALGEBRAIC  METHOD     [CHAP.  XII 

point  where  there  is  no  concentrated  load  or  reaction 
the  corresponding  value  of  x  may  be  used  in  the  shear 
equation,  with  V  equal  to  zero  and  the  value  of  the 
constant  Vi  determined.  However,  this  substitution  will 
seldom  be  necessary,  as  the  value  of  V\  will  usually  be 
determined  by  other  methods. 

(b)  The  section  of  zero  bending  moment  will  be  at  the 
free  ends  of  beams,  as  at  A,  Fig.  92,  and  at  the  ends 
supported  without  restraint;  also  at  the  points  of  inflec- 
tion for  overhanging,  continuous,  and  restrained  beams, 
as  at  /,  Fig.  92,  but  the  inflection  points  in  such  beams 
cannot  be  obtained  by  inspection. 

(c)  The  section  of  zero  slope  is  at  the  horizontal  por- 
tion of  the  beam,  as  at  D,  Fig.  92.     For  symmetrical 
beams   carrying   symmetrical   loads   the   beam   is   hori- 
zontal at  the  axis  of  symmetry   (at  the  center).     By 
definition  beams  with  fixed  ends  are  horizontal  at  the 
fixed  ends. 

(d)  For  the  axes  chosen  the  section  of  zero  deflection 
is  at  the  supports.     For  overhanging   and   continuous 
beams  there  may  be  one  or  two  positions  in  a  span 
where  the  deflection  is  zero,  but  these  points  cannot  be 
determined  by  inspection. 

120.  ESSENTIAL  QUANTITIES  TO  BE  KNOWN  ABOUT 

BEAMS.  In  all  kinds  of  beams  the  important  things  to 
be  obtained  are  the  position  and  magnitude  of  the 
maximum  stresses  and  the  maximum  deflection.  For 
overhanging,  continuous,  and  fixed  beams  the  inflection 
points  need  to  be  found.  When  the  maximum  vertical 
shear  is  determined,  the  maximum  shearing  stress  is 

y 
then  obtained  by  use  of  the  shear  formula  J  =  rr '     When 

the  maximum  bending  moment  is  found  the  maximum 
fiber  stress  developed  may  be  obtained  by  use  of  the 


EXAM.]       ELASTIC  CURVE  —  ALGEBRAIC  METHOD          197 


moment  formula/  =  —=-• 


These  formulas  may  be  used 


to  determine  the  safe  load  for  a  given  beam,  and  also 
to  design  beams.  Building  specifications  usually  state 
that  the  maximum  deflection  shall  not  exceed  a  given 
amount,  therefore  it  is  necessary  to  be  able  to  determine 
the  maximum  deflection  for  beams. 


EXAMPLES 

i.  Deduce  the  equation  of  the  elastic  curve  and  the  maximum 
deflection  for  a  cantilever  beam  with  a  uniform  load  of  w  pounds 
per  lineal  unit.     See  Fig.  93. 
The  load  per  lineal  unit  is  —  w. 


FIG.  93. 


When    x  =  l,   a  =  o, 
/.     o  =  — 


When   x  —  o,    V  =  o,    /. 
M 


When    x  =  o,   1,1  =  o,    . 


Fx  =  o.     (Zero  shear), 
=-  Cwxdx  +  Ml  =  -—  *  + 

J  2 

x  =  o.     (Zero  moment), 


g£j  +  «i   and  «i=^gj--    (Zero  slope), 


'*'     a          6EI+ 6EI' 


198     ELASTIC  CURVE  —  ALGEBRAIC  METHOD    [CHAP.  XII 


/,  C  wxfdx  .     C 

•AH-*  — J  -6£7+J 


When    x  =  /,   y  =  o, 


6EI 


wl* 


0  = 


and      yl  =  -  -—•    (Zero  deflection), 
8  hii 


SEI 


(Elastic  curve). 


The  maximum  deflection  is  at  the  end,  where  x  =  o,  and  is 
w^__        WP_ 
&EI          SEI' 

2.  Deduce  the  equation  of  the  elastic  curve  and  the  maximum 
deflection  for  a  simple  beam  with  a  uniform  load.     See  Fig.  94. 
The  load  per  lineal  unit  is  —  w. 


o  '* 


FIG.  94. 


When  s  =  o  the  vertical  shear  is  equal  to  the  left  reaction,  which 

.    wl 


—  \  wxdx  +  \  - 


22 

When  oj  =  o,   M  =  o,    .'.    Jkfj  =  o.     (Zero  moment) . 


EXAM.]       ELASTIC  CURVE  —  ALGEBRAIC  METHOD          199 


The  beam  is  horizontal  at  the  center,  hence 

w? 


when  x  =  -,   a  =  o. 
2 


^  f  wx?dx  ~  m 


_       wx*         wlx*         wPx 
~  24EI      i2EI~  24EI     yi' 

When  x  =  o  the   deflection  is  zero,  or   y  =  o,     /.  y{  =  o.     (Zero 
deflection), 

wx*        wlx? 


The  maximum  deflection  occurs  at  the  center  and  is  obtained  by 
letting  x  =  -  in  the  equation  for  y,  and  it  is  found  to  be 

_^  ^  =     J.  WL 
384  El         384  El  ' 

3.  Deduce  the  equation  of  the  elastic  curve,  and  the  value  of 
the  maximum  deflection  for  a  simple  beam  with  a  concentrated 


FIG.  95. 

load  at  the  distance  kl  from  the  left  support  in  which  k  is  a  fraction. 
See  Fig.  95. 

Ri  =  W(i-  k}. 

The  load  curve  is  at  zero.    To  the  left  of  the  load, 
V  =  W(i-k}. 

U  =  f  W(i  -  k}  dx  +  M,  =  W  (i  -  k)  x  +  Mi, 

when        x  =  o,  M  =  o,    /.  M :  =  o, 

...     M  =  W(i-k)x.  (b) 

(c) 


200    ELASTIC  CURVE  —  ALGEBRAIC  METHOD    [CHAP.  XII 

The  beam  is  horizontal  between  the  load  and  the  center  but  for 
what  value  of  x  it  is  not  known,  hence  we  must  let  on  remain  in 
the  equations  till  enough  conditions  are  obtained  to  determine  its 
value. 

y  =  -gj  §  (i  -  k)  °^  dx  +  J"  aidx  +  yi, 


When  x  =  o,  y  =  o,    .:  y^  =  o, 

W 


(d) 


This  is  all  that  is  known  about  the  equations  to  the  left  of  the 
load,  hence  we  must  use  those  to  the  right  of  the  load. 
To  the  right  of  the  load, 

V=-Wk.  (i) 

M  =  -  f  Wkdx  +  M  /  =  -  Wkx  +  MS. 

When  x  =  l,  M  =  o,   /.  MS  =  Wkl,  and  M  =  Wk  (I  -  x)        (2) 

Wkrn  / 

«  =  —  J  (l-xJdx  +  aS, 


The  value  of  a.  for  this  curve  is  not  known  for  any  value  of 
x.  Hence  «/  must  be  kept  in  the  equation  till  its  value  can  be 
determined. 


TTTl  7  /  I*     &V  It 

When  x  =  /,   y  =  o,    .:  yt  =  --  —  —  ai  I, 

3  El 

,        Wkl3 


EXAM.]      ELASTIC  CURVE  —  ALGEBRAIC  METHOD          201 

To  determine  «i  and  «/  it  is  known  that  the  slope  of  both 
portions  of  the  elastic  curve  at  the  load  are  the  same.  Therefore 
(c)  and  (3)  are  equal  when  x  equals  kl. 

W,        .v*1/1 


This  gives  one  relation  between  <xi  and  «/.  To  obtain  another 
relation  between  them,  it  is  known  that  the  deflection  of  both 
portions  of  the  elastic  curve  are  the  same  at  the  load.  Therefore 
(d)  and  (4)  are  equal  when  x  equals  kl. 

w         m* 


Solving  equations  (A)  and  (B)  for  «i  and  a/, 
Wl2 


6EI 

WP 


(2 


To  the  left  of  the  load,  then, 


To  the  right  of  the  load, 

Wkx3  ,   Wklx*      WPx,    i.^ 
""6E/4"^E7""6E7(2/      k) 

The  value  of  x  for  the  maximum  deflection  is  obtained  by  equating 

a  to  zero  and  solving  for  x.    If  k  is  greater  than  -  ,  the  value  of  x 

2 

is  found  to  be 


3  (i  ~  *) 

This  value  of  x  substituted  in  the  expression  for  y  gives  the 
maximum  deflection  to  be 


202     ELASTIC  CURVE  —  ALGEBRAIC  METHOD    [CHAP.  XII 


4.  Deduce  the  equation  of  the  elastic  curve  and  determine 
the  value  of  the  maximum  deflection  and  the  maximum  moment, 
and  locate  the  inflection  points  for  a  beam  fixed  at  both  ends  with 
a  concentrated  load  at  the  center.  See  Fig.  96. 

From  symmetry  both 
reactions  are  equal. 
The  load  curve  is  zero. 
To  the  left  of  the  load, 


FIG.  96. 


The  bending  moment  at  the  wall,  which  is  the  restraining 
moment  and  keeps  the  beam  horizontal  at  that  point,  is  not 
known  at  the  start.  The  position  of  zero  moment  is  not  known 
either,  so  Mi  must  be  retained  at  the  present. 

Wx2 


• 
4  El       El 

By  definition  of  restrained  beam  the  slope  at  the  wall  is  zero, 
therefore  when  x  =  o,  a  =  o,  and  ai  =  o.  From  symmetry  it 
is  seen  that  the  beam  is  also  horizontal  at  the  center,  therefore 


when  x  =  -  ,  a  —  o. 
2 


o  = 


Wl2        MJ 


T 
i6EI      2  El 


and 


Wl 
— 

8 


_  Wx      Wl 


Wxz 
4EI 

Wx3 

12  El      i6EI 


.Wlx 
'  8EI' 
Wlx2 


When 


o,    y  =  o,     .'.   ; 
Wx*         Wlx2 


12  El      i6EI 


EXAM.]     ELASTIC  CURVE  —  ALGEBRAIC  METHOD 


203 


The  maximum  deflection  occurs  at  the  center  and  is  found  to  be 

Wl3         Wl3  WP 


96  El      6^EI         192  El 
Similar  equations  can  be  deduced  for  the  right  half  of  the  beam. 

Wl 

The  bending  moment  at  the  wall  is  --  —  ,  and  at  the  middle 

8 

of  the  beam  it  is  +  —  • 

o 

The  inflection  point  is  the  point  at  which  the  bending  moment 
is  zero,  and  is  found  by  equating  M  to  zero  and  solving  for  the 
corresponding  value  of  x,  which  may  be  called  x\. 


Wx,      Wl 


5.    Determine  the  left  reaction,  the  maximum  bending  moment, 
and  the  equation  of  the  elastic  curve,  and  locate  the  inflection 


FIG.  97. 

point  for  a  beam  fixed  at  one  end  and  supported  at  the  other 
end,  carrying  a  uniform  load.     See  Fig.  97. 
The  load  per  lineal  unit  is  —  wt 


Since  the  reactions  cannot  be  determined  at  the  start,  the  value 
of  Vi  cannot  be  determined  at  first.  The  left  reaction  is  less 
than  it  would  be  if  the  beam  were  not  restrained  at  the  right  end. 


When 


2 

M  =  o,  /. 
Wx3  V&* 
6E7  2EI 


Afi-o, 


204     ELASTIC  CURVE  —  ALGEBRAIC  METHOD    [CHAP.  XII 

When  *  =  /,   «  =  o. 

wP_  <Vjtf  wl*        Fi/2 

6EI      2EI  6EI      2EI' 

wxz       V&\     wl3       Fi/2 
6EI      2EI      6EI      2EI' 


VJx 
24EI  "*"  6  El      6  El      2EI~t  yi' 


When  x  =  o,   y  =  o,    /.  y\  =  o. 

Also  when        x  =  I,   y  =  o, 

w/4        FiP        wl* 


2EI' 


wx*         wlo? 
~  - 


i6El 

The  maximum  positive  moment  is  at  the  distance  f  /  from 
the  left  support,  and  is  Mm  =  T|B  wlz.  The  maximum  negative 
moment  is  at  the  wall  and  is  Mm  =  —  |  wl2.  The  inflection  point 
is  at  II  from  the  left  end.  The  point  of  maximum  deflection 
occurs  where  the  slope  is  zero  between  the  supports. 

6.  Determine  the  relation  between  the  bending  moments  over 
the  three  supports  of  two  consecutive  spans  of  a  continuous 
beam  carrying  uniform  loads  on  both  spans.  See  Fig.  98. 


FIG.  98. 

Using  the  relations  deduced  in  Art.  116  we  have  for  the  first 
span: 

The  load  per  unit  of  length  is  -  Wi. 

V  =  -  W&  +  Fi.     (No  relations  yet  to  determine  Fj).  (a) 


M=—    -=r  +  Fj#+  MI.     (No  relations  yet  to  determine  M  \).    (b) 


EXAM.]     ELASTIC  CURVE  —  ALGEBRAIC  METHOD  265 


a  = 


+         +         +ai.     (No  relations  yet  to  determine  «,)  .  (c) 


6  £/     2  El      El 


MVXZ    .  xjv 


When  x  =  o,   y  =  o.      /.  y^  =  o. 

When  of  =  lit  y  =  o, 


6EI      2EI 

By  the  same  process  we  obtain  equations  for  the  second  span. 

For  the  second  span, 

The  load  per  unit  of  length  is  —  w2. 

V  =  —  w2x  +  F2.     (No  relations  yet  to  determine  F2).  (i) 

M  =  —  ^-  +  V2x  +  MI.     (No  relations  to  determine  M2).       (2) 

w  x3      V  x2     M  x 
a=-    ~  +  -       +  -±-  +«2.    (No  relations  to  determine  a,)  .  (3) 


When  x  =  o,   y  =  o,    /.  yz  =  o 

Also  when  x  =  12,   y  =  o,  and 

2/23        F2/22 


24  £/  6£/  2.E/ 
It  is  known  that  the  slope  of  each  portion  of  the  elastic  curve 
is  the  same  at  the  middle  support.  By  letting  x  =  l^  in  equation 
(c),  and  letting  x  =  o  in  equation  (3),  and  equating  (c)  to  (3), 
and  remembering  the  value  of  V±  and  F2  from  Art  113,  there 
results  the  theorem  of  three  moments, 

MA  +  2  M  2  (/!  +  /,)  +  MA  =  -  H^!  -  ^  . 

4  4 

7.  What  is  the  maximum  deflection  of  an  8-inch,  1  8-pound 
cantilever  I-beam  10  feet  long  carrying  a  load  of  1800  pounds 
concentrated  at  the  end? 

Wl3       1800X120X120X120 
A  =  —  —  .  =  —  --  —  =  0.608  inch. 

3  El        3  X  30,000,000  X  56.9 


206     ELASTIC  CURVE  —  ALGEBRAIC  METHOD    [CHAP.  XII 


PROBLEMS 

i.  For  a  simple  beam  with  a  concentrated  load  at  the  center 
show  that  the  equation  of  the  elastic  curve  to  the  left  of  the 
center  is 


and  show  that  the  maximum  deflection  is 


48  EI 

2.  Deduce  the  equation  of  the  elastic  curve  for  a  cantilever 
beam  carrying  a  load  W  concentrated  at  the  end  and  a  uniform 
load  w  per  lineal  unit.    Also  determine  the  maximum  deflection. 

Wl*        wV 
Ans,  A  —  _-_. 

3.  Deduce  the  value  of  the  maximum  deflection  for  a  simple 
beam  carrying  a  load  W  concentrated  at  the  center  and  a  uniform 

load  w  per  lineal  unit.  Wl3          «:  wl* 

Ans.  A  =  —  ---  -  -- 

48  £7      384^7 

4.  (a)  Deduce  the  equation  of  the  elastic  curve  for  a  beam 
fixed  at  both  ends  carrying  a  uniform  load,     (b)  Determine  the 
value  of  the  maximum  bending  moment,     (c)  Determine  the  value 
of  the  maximum  deflection,     (d)  Locate  the  inflection  points. 

Ans.  (a) 


/UN       Ttf  Wl* 

(b)   M  m  =  --  , 


(d)       x,  = 

5.   For  a  beam  fixed  at  both  ends  and  carrying  a  concentrated 
load  at  the  distance  kl  from  the  left  support  show  that: 
The  left  reaction  is  ^  =  W  (i  -  3  k2  +  2  yfc3). 
The  moment  at  the  left  support  is  Ml  =  -  Wlk  (i  —  2  k  +  k2). 
The  moment  under  the  load  is  M3  =  Wltf  (2  —  4  k  +  2  £2). 
Also  determine  the  value  of  the  deflection  y. 


PROB.]       ELASTIC  CURVE  —  ALGEBRAIC  METHOD  207 

6.   For  a  beam  fixed  at  one  end  and  supported  at  the  other, 
and  carrying  a  concentrated  load  at  the  center,  show  that: 

The  left  reaction  is  ^  W. 

The  moment  at  the  wall  is  —  T\  Wl. 
.    The  moment  at  the  load  is  -fa  Wl. 

The  values  of  y  are: 


O 

The  inflection  point  is  —  /  from  the  free  support. 

7.  For  a  beam  fixed  at  the  right  end  and  supported  at  the  left 
carrying  a  concentrated  load  of  W  at  the  distance  kl  from  the 
left  support  show  that: 

W  A 

The  reaction  at  the  left  end  is  —  (2  -  3  k  +  /?). 

Wl 
The  moment  at  the  wall  is  -  —  -  (k  -  k3). 

2 
Wlk 

The  moment  at  the  load  is  -  -  (2  -  3  k  +  k3}. 

8.  For  equal  spans  and  equal  uniform  loads  on  all  spans  show 
that  the  theorem  of  three  moments  reduces  to 


9.   Solve  the  problems  at  the  end  of  Chapters  IX  and  X. 


CHAPTER  XIII 
SECONDARY  STRESSES 

121.  HORIZONTAL  SHEAR  IN  BEAMS.    When   one 

board  is  placed  on  top  of  another  one  and  the  two  are  then 
used  as  a  beam  the  upper  board  will  slip  over  the  lower  one 
in  one  direction  at  one  end  and  in  the  opposite  direction 
at  the  other  end.  To  prevent  this  motion  and  to  make 
the  beam  stronger  the  boards  may  be  nailed  together, 
the  nails  taking  shear.  In  all  beams  there  is  the  tendency 


«r  1/1  Z\ 

A  IB 


D, 


ShXiSl 


Vl  Zl 


FIG.  99. 

of  the  upper  part  to  slip  past  the  lower  part  along  any 
horizontal  plane,  a  horizontal  shearing  stress  thus  being 
produced. 

It  is  the  object  of  this  article  to  show  that  the  vertical 
shearing  unit-stress  and  the  horizontal  shearing  unit- 
stress  at  any  point  in  a  beam  are  the  same.  Proof:  In  a 

208 


ART.  122]  SHEARING  UNIT-STRESSES  2OQ 

beam  let  a  small  parallelepiped  ABDC,  of  dimensions  %i, 
3?i,  Zi,  be  imagined  cut  from  the  beam.  Neglecting  the 
effect  upon  the  shearing  stress  of  any  load  for  the  element 
of  length  xit  the  vertical  shearing  unit-stress  will  be  the 
same  on  both  vertical  faces,  and  the  horizontal  shearing 
unit-stress  will  be  practically  the  same  on  the  top  and 
bottom  faces.  The  forces  acting  upon  the  element  as 
shown  in  Fig.  99  are  svy\z\  on  the  vertical  faces,  and 
ShXiZi  on  the  top  and  bottom  faces,  in  which  sv  is  the 
vertical  shearing  unit-stress,  and  sh  is  the  horizontal 
shearing  unit-stress.  By  taking  moments  about  any 
point,  as  A,  the  relation  between  the  two  unit-stresses 
is  deduced. 

2  MA  =  ShX&iyi  —  svyiZiXi  =  o, 

.'.     sh  =  sv. 

122.  THE  MAGNITUDE  OF  THE  HORIZONTAL  AND 
VERTICAL  SHEARING  UNIT-STRESSES  AT  A  POINT.    In 

Art.  63  it  was  assumed  that  the  maximum  vertical  shear- 
ing unit-stress  for  a  section  is  greater  than  the  average,  and 
values  of  the  ratio  between  the  two  were  given  for  several 
standard  sections.  To  determine  those  ratios  the  value 
of  the  horizontal  shearing  unit-stress  must  be  deduced. 
The  expression  for  the  horizontal  shearing  unit-stress  will 
now  be  deduced.  Let  Fig.  100  (a)  represent  a  beam  with  a 
portion  A  BCD  imagined  cut  from  the  beam.  The  stresses 
on  the  fibers  of  the  section  AD  in  general  will  not  be  equal 
to  the  stresses  on  the  fibers  of  the  section  BC,  because 
the  bending  moments  at  the  two  sections  are  usually 
different.  If  the  bending  moment  at  the  section  AD  is 
less  than  that  at  the  section  BC,  the  resultant  HI  of  the 
stresses  acting  upon  the  face  AD  is  less  than  H2,  which 
is  the  resultant  of  the  stresses  acting  upon  the  face  BC. 
To  maintain  equilibrium  a  horizontal  shearing  force 
snub  must  act  upon  the  face  CD,  sh  is  the  horizontal 


2IO 


SECONDARY  STRESSES 


[CHAP.  XIII 


shearing  unit-stress,  u  is  an  element  of  length,  AB,  and 
b  is  the  thickness  of  the  beam. 


HrV 

D 

c 

—  /-H2 

1 

1 

; 

/ 

IA/ 

/             Neutral 

surface 

l«) 

FIG.  ioo. 

By  Art.  71  #1  is  equal  to  the  area  of  the  section  above 
the  plane  multiplied  by  the  unit-stress  on  the  centroid 
of  that  area. 


Likewise 


/i  is  the  unit-stress  developed  on  the  outside  fiber  at  the 
section  AD,  and  /2  is  the  unit-stress  on  the  outside  fiber 

/I  JVI     1 

at  the  section  BC.     From  the  moment  formula,  —  =  -y- 

and  —  =  ~~-     For  equilibrium  of  the  element: 
c        J. 

H2  —  HI  =  shub, 
f    lA'yr  -- 


—  MI  At-,  , 

—  -A'y'  =  shub, 


A' 


*  = 


ART.  122] 


SHEARING  UNIT-STRESSES 


M2-  Ml 


is  the  rate  of  change  of  the  bending  moment, 


and  this  equals  V  when  u  is  made  indefinitely  small. 
A'y'  is  zero  for  y  equal  to  c  and  is  a  maximum  for  y 
equal  to  zero.  Therefore  the  shearing  unit-stress  is 
zero  at  the  outside  fiber  of  beams,  and  is  greatest  at  the 
neutral  surface. 

The  maximum  shearing  stress,  both  horizontal  and 
vertical,  developed  in  a  rectangular  beam  of  breadth  b 
and  depth  d  where  V  is  the  vertical  shearing  force,  is 

"•*  m     "nrT    *i       r          ^~7^"     ^>  '      S^>  7       7     "~~     '  A 

Ib  bd3b       2       4      2bd      2  A 

12 

This  is  one-half  greater  than  the  average  vertical  shear- 
ing stress. 

For  circular  sections  the  maximum  shearing  stress  is 
V 


64 

This   is    one-third    greater    than    the    average    vertical 
shearing  stress. 

For  built-up  and  I -sections  the  maximum  shearing 
stress  is  approximately  equal  to  that  obtained  by  divid- 
ing the  vertical  shear  by  the  area  of  the  web  AI. 


Rectangle 


I-Section 


The  variation  in  the  intensity  of  the  shearing  unit- 
stress  for  various  sections  is  shown  by  the  diagrams  of 
Fig.  1 01. 


212  SECONDARY  STRESSES  [CHAP.  XIII 

Because  of  their  small  strength  in  shear  parallel  to  the 
grain,  timber  beams  frequently  fail  by  shearing  along 
the  neutral  surface.  Beams  should  always  be  investi- 
gated for  the  maximum  shearing  stress  developed. 

123.  PLATE  GIRDERS,  FIRST  METHOD.    One  method 

of  the  design  of  plate  girders  is  to  consider  all  plates 
and  angles  acting  as  a  homogeneous  beam.  The  girder 
may  then  be  designed  by  the  use  of  the  moment  and 
shear  formulas,  and  the  pitch  of  the  rivets  can  be 

determined  by  the  use  of  the  formula  for  horizontal 

y 
shear.     In  Sh  =  jrA'y',  sh  is  the  stress  developed  upon  a 

unit  of  area  of  the  horizontal  plane.  Multiplying  this 
by  b  gives  the  total  force  which  would  be  transmitted 
from  the  upper  section  to  the  lower  one  in  one  unit  of 
length  of  the  girder.  For  built-up  sections  the  stress 
must  be  transmitted  from  the  upper  plates  to  those  next 
below,  through  the  rivets  connecting  the  plates.  If  p  is 
the  pitch  of  the  rivets,  and  there  are  n  rivets  in  the  dis- 
tance p,  the  force  that  each  rivet  must  carry  will  be 


. 
n         In 

The  greatest  number  of  rivets  will  be  required  where  the 
product  VA'  is  greatest. 

124.  PLATE  GIRDERS,  SECOND  METHOD.   Another 

method  of  design  for  plate  girders  is  to  assume  that  all 
the  tensile  and  compressive  stresses  are  taken  by  the 
flanges,  and  that  the  stress  is  uniform  over  the  section 
of  the  flanges,  and  that  the  shear  is  taken  by  the  web. 
The  stresses  calculated  in  this  way  are  probably  a  little 
in  excess  of  those  actually  developed,  but  the  error  is  on 
the  side  of  safety. 


ART.  124]      PLATE    GIRDERS,  SECOND   METHOD 


213 


FIG.  102. 


Let  Fig.  1 02  be  the  cross  section  of  a  girder,  A  the 
effective  area  of  one  flange,  d  the  distance  between  the 
centroids  of  the  flanges,  and  /  the 
unit-stress  developed  in  the  flanges, 
then  the  compressive  and  tensile 
forces  equal  Af  and  the  resultants 
act  at  the  centroids  of  the  flanges. 
The  moment  of  these  internal 
stresses  resists  the  bending  mo- 
ment due  to  the  external  forces. 
For  equilibrium,  then, 

M  =  Afd, 

where  M  is  the  bending  moment  and  Afd  is  the  resisting 

moment. 

The  bending  moment  increases  toward  the  center  of 

the  span,   and  to  increase  the  resisting  moment  with 

the  same  unit-stress  the 
area  A  is  increased  by  add- 
ing cover  plates  in  the 
center.  The  pitch  p  of 
the  rivets  connecting  the 
flanges  to  the  web  may  now 
be  found.  (See  Fig.  103.) 
The  change  in  the  stress 


FIG.  103. 


in  the  flanges  between  any 
two  sections  must  be  trans- 

mitted through  the  rivets  to  the  web.     This  difference 

in  a  unit  of  length  is 


-  Mi 


V 

d  d' 

since  Mt  —  MI  is  the  rate  of  change  of  the  bending 
moment,  as  the  distance  between  the  sections  was  taken 
as  unity.  If  the  rate  of  change  of  the  bending  moment, 


214  SECONDARY  STRESSES  [CHAP.  XIII 


or  F,  is  constant,  the  change  for  the  distance  p  is  p  -j- 

If  there  are  n  rivets  in  the  pitch  p,  and  R  is  the  allowable 
force  one  rivet  will  transmit, 


_nRd 
P" 


125.  COMBINED  FLEXURE  AND  TENSION  OR  COM- 
PRESSION. When  a  beam  is  subject  to  axial  loads  in 
connection  with  the  flexural  loads  the  maximum  stress 
developed  may  be  considered  as  made  up  of  two  parts  — 

that  due  to  bending  and 

J  Load  L  —    that  due  to  the  axial  load. 

The   axial  load   increases 

Ri  R!S       the  tensile  oj  compressive 

B  &        F  stress  due  to  the  bending. 

"/  Fig.  104  (b)  is  a  free-body 

diagram  of  a  portion  of  the 


c  D     E.  beam  under  a  compression 

(b)  load  P.     If  the  deflection 

FIG.  104.  of  the  beam  is  small  the 

moment  due  to  P  may  be 
neglected.   If  M  is  the  bending  moment  due  to  the  flexural 

loads,  A  the  sectional  area,  and  -  the  section  modulus, 

0 

Me 
the    maximum    flexural    stress    developed    is    /i  =  -j- 

(indicated  by  AB  in  compression).     The  compressive 

p 
unit-stress  due  to  the  axial  load  is  /2  =  -r  (indicated  by 

A 

FA).     It  is  seen  that  the  maximum  compressive  stress 
is  developed  in  the  most  remote  fiber  in  compression 


ART.  125]       COMBINED  FLEXURE  AND  TENSION,  215 

and  equals  the  sum  of   the   flexural   and  direct  stres"s 
and  is 

Me      P 


If  P  is  a  tension  load  the  maximum  tensile  stress  is 
of  the  same  form  as  that  given  for  compression. 

In  longer  beams,  where  the  deflection  is  appreciable, 
there  is  an  additional  moment  M'  due  to  the  load  P. 

The  moment  is  decreased  if 

P    /-"-^ 

P  is  tension  and   increased 

if  P  is   compression.      (See 

Fig.  105.)     If  Ai  is  the  max- 

imum  deflection  due  to  both  pIGi  I05 

the    transverse    and    longi- 

tudinal loads  the  moment  of  P  is  PAi,  and  the  total 

moment  is 

M  ±  M'  =  M  ±  PAi. 

,       MIC      (M  ±  PAi)c 

h=  ~T         ~T 

The  plus  sign  is  for  a  compression  load  and  the  minus 
is  for  a  tension  load.  In  order  to  find  the  stress  AB,or 
/i  due  to  the  moment  of  both  loads,  Ai  must  be  expressed 
in  terms  of  /i,  the  maximum  stress  for  the  deflection  AI. 

From  Art.  107,    Ai  =  ^j~  - 

pJlLC 

Me 


from  which  fi  —  - 


p 
To  this  add  the  direct  stress  -j  due  to  the  axial  load. 


2l6  SECONDARY  STRESSES  [CHAP.  XIII 

Then  the  maximum  stress  developed  is  found  to  be 

f    f  .P     Me  i      i      \  .  p 

J-Ji  +  A-~T 


7 ! \ 

•  -  s 


The  minus  sign  is  used  for  P  in  compression  and  the 
plus  for  P  in  tension. 

126.  COMBINED  SHEARING  STRESSES  AND  TENSILE 

OR  COMPRESSIVE  STRESSES.     Let  Fig.  106  represent  a 
small  portion  of  a  beam  where  the 
c  ;<  x-*~~    >[  B    known  unit-stresses  are  s  in  shear 

~  anc^  /  m  tension  or  compression,  5 


fvz  \\£^^'^ruz          and  /  being  at  right  angles  to  each 
ATJfcu*^  other.     Along  all  diagonal  planes, 

FIG.  106.  as  AB,  there  are  normal  and  tan- 

gential components  of  the  stresses. 

sp  is  the  shearing  unit-stress  along  the  plane  and  /„  is 
the  tensile  or  compressive  unit-stress  normal  to  the  plane. 
In  more  advanced  texts  it  is  shown  that  the  value  of  0 

to  give  the  maximum  sp  is  such  that  tan  2  <$>  =  — ,  and 

2  S 

the  corresponding  maximum  shearing  stress  is 


S-o     — 


The  value  of  <£  to  give  the  maximum  fn  is  such  that 
cot  20=—  —  ,  and  the  maximum  tensile  stress  is 


In  the  latter  equation  the  maximum  stress  will  be 
obtained  with  the  plus  sign,  and  if  fn  is  a  tensile  stress, 
the  maximum  /n  will  be  a  tensile  stress.  If  /  is  a  com- 
pressive stress  the  maximum  /„  will  be  a  compressive 


EXAM.]  SECONDARY   STRESSES'  217 

stress.  If  the  minus  sign  is  used  before  the  radical 
the  resulting  fn  will  be  negative,  which  indicates  that 
it  is  of  opposite  sign  from/,  i.e.,  /„  is  compression  for/ 
tension,  and/n  is  tension  for/  compression. 


EXAMPLES. 

i.  Determine  the  maximum  horizontal  shearing  unit-stress 
in  a  timber  beam  8  inches  by  14  inches  under  a  load  of  24,000 
pounds  applied  at  the  third  points. 

Ri  =  jR2  =  1  2,  ooo  pounds. 


2.  Compute  the  pitch  of  |-inch  rivets  for  a  plate  girder  of 
72-ft.  span  and  7  feet  2  inches  deep,  the  cover  plates  being  14  inches 
wide  with  a  total  thickness  of  |  inch  at  the  center,  being  connected 
to  angles  |  inch  thick.  (NOTE:  This  girder  was  designed  to  carry 
a  live  train  load  together  with  the  weight  of  the  tracks  and  the 
girder.) 

The  maximum  vertical  shear  at  the  ends  was  found  to  be  137,000 
pounds;  at  5  feet  from  the  ends,  121,000  pounds;  at  10  feet  from 
the  ends,  102,000  pounds,  etc. 

Each  rivet  will  carry  .601  X  8000  =4810  pounds  in  shear,  and 
f  X  |  X  18,000  =  5900  pounds  in  bearing.  The  shear  governs  in 
this  case. 

Taking  2  rivets  in  the  pitch  p, 

2  X  4810  X  86 

p  =  -  —     -  =  6  inches  at  the  ends. 
137,000 

p  =  6.8  inches  at  5  feet, 

p  =  8.05  inches  at  10  feet,  etc. 

For  concentration  of  the  loads  on  the  girder  the  maximum 
allowable  pitch  would  be  about  6  inches. 


2l8  SECONDARY  STRESSES  [CHAP.  XIII 

3.  A  timber  8  inches  by  10  inches  is  used  as  a  simple  beam  of 
i2-ft.  span  to  carry  a  uniform  load  of  4000  pounds  and  end  com- 
pression loads  of  40,000  pounds.  What  is  the  maximum  stress 
developed? 

By  assuming  the  deflection  negligible, 

,. P      Me  _  40,000      4000  X  144  X  5  X  12 

-~  A      ~T         80          8X8X  loXioX  10' 
=  500  +  540  =  1040  pounds  per  square  inch. 

By  use  of  the  formula  assuming  the  deflection  not  negligible, 


=  500  +  540 


8  X  40,000  X  I442  X  3   I 
76.8  X  1,500,000  X  2000' 

=  500+  54o( ^-37- )  =  5°°+  -^~z 

V i -.08647  •9I36 

=  500  +  590  =  1090  pounds  per  square  inch. 

Thus  it  is  seen  that  the  moment  of  the  axial  load  about  the 
central  section  increases  the  stress  about  5  per  cent. 

4.  A  bolt  i  inch  in  diameter  is  subjected  to  a  tension  of  3000 
pounds  and  at  the  same  time  to  a  cross  shear  of  5000  pounds. 
Determine  the  maximum  tensile  and  shearing  unit-stresses. 

5  =  5000  4-  .7854  =  6370  pounds  per  square  inch. 
/  =  3000  -5-  .7854  =  3820  pounds  per  square  inch. 

By  substitution  in  the  formulas  for  the  maximum  tensile  and 
shearing  stresses, 


i/ 
V 


fn  =          +  i63702  +         -  =  8560  Ib.  per  sq.  in.,  tension. 


,-v/ 


6370*  -f- =  6650  Ib.  per  sq.  in.,  shear. 


PROB.]  SECONDARY   STRESSES  219 

PROBLEMS 

1.  A  simple  rectangular  timber  beam  8  inches  by  12  inches  and 
of   lo-ft.  span  carries  a  uniform  load  of  2000  pounds  per  foot. 
Determine   the   horizontal   shearing  unit-stress  at  the  following 
points:  (a)  At  the  neutral  surface  over  a  support,     (b)  3  inches 
from  the  neutral  surface  over  a  support,     (c)  4  inches  from  the 
neutral  surface  at  a  quarter  point. 

2.  What  is  the  maximum  shearing  unit-stress  developed  in  an 
I-beam  of  the  largest  standard  section  which  carries  a  uniform  load 
over  a  span  of  20  feet  if  the  maximum  fiber  stress  does  not 
exceed  16,000  pounds  per  square  inch? 

3.  A  girder  of  55^-ft.  span  is  built  up   of   £-inch  by  4  feet 
lo-inch  web  plate,  four  5-inch  by  6-inch  by  f-inch  angles,  with  the 
5-inch  leg  riveted  to  the  web,  and  four  cover  plates  at  the  quarter 
points  14  inches  by  \  inch.     The  rivets  are  |  inch  in  diameter  and 
spaced  3  inches  apart,  and  there  are  two  rows  in  each  5-inch  leg. 
Determine  the  maximum  shearing  and  bearing  unit-stresses  that 
would  probably  come  on  the  rivets.     (Section  is  similar  to  that 
shown  in  Fig.  102  with  cover  plates  added.     Girder  is  to  carry  a 
train  load.) 

4.  Determine    the   maximum   stress   developed   in  a   6-inch, 
15-pound  I-beam  of  6-ft.  span  with  both  ends  fixed,  carrying  a 
uniform  load  of  8  tons  and  tension  loads  at  the  ends  of  6  tons. 

5.  What  will  be  the  maximum  fiber  stress  developed  in  a  simple 
timber  beam  6  inches  by  8  inches  of   8-ft.  span,  with  a  concen- 
trated load  of  1500  pounds  at  the  center  and  end  compression 
loads  of  10,000  pounds? 

6.  A  i2-inch,  4O-pound  I-beam  of  6-ft.  span  carries  a  uniform 
load  of  1 200  pounds  per  foot,  and  is  subjected  to  an  axial  compres- 
sion of  60,000  pounds.     Find  the  maximum  stress  developed. 

7.  Find  the  size  of  a  square  maple  simple  beam  for  a  simple 
span  of  12  feet  to  carry  a  load  of  500  pounds  at  the  middle,  when 
it  is  also  subjected  to  an  axial  compression  of  2000  pounds. 

8.  A  bar  of  iron  is  under  a  direct  tensile  stress  of  4000  pounds 
per  square  inch  and  a  shearing  stress  of  3500  pounds  per  square 
inch.     Find  the  maximum  tensile  and  shearing  unit-stresses. 

9.  Design  a  white  oak  beam  with  both  ends  fixed,  for  a  span  of 


220  SECONDARY   STRESSES  [CHAP.  XIII 

12  feet,  which  is  to  carry  a  concentrated  load  of  4  tons  at  the 
center  and  a  tension  load  of  5  tons. 

10.  What  I-beam  would  be  required  for  the  loading  given  in 
Problem  No.  6  if  the  unit-stress  is  not  to  exceed  16,000  pounds 
per  square  inch? 

n.  What  will  be  the  maximum  shearing  and  tensile  unit- 
stresses  developed  in  a  |-inch  bolt  if  it  is  subjected  to  a  tension 
load  of  5000  pounds  and  a  cross  shearing  load  of  5000  pounds? 


CHAPTER   XIV 


COLUMNS  AND   STRUTS 

127.  DISCUSSION.     The   terms    columns   and    struts 
are  usually  applied  to  prismatic  members  designed  to 
carry  compression  loads  when  the  length 

has  an  effect  on  the  strength  of  the 
member.  For  short  compression  mem- 
bers, lateral  deflection  is  inappreciable 
under  load,  while  for  longer  ones  (i.e. 
columns)  it  may  be  of  consequence. 
Long  columns  will  not  carry  so  great 
loads  as  shorter  ones  of  the  same  mate- 
rial and  section,  since  the  lateral  bend- 
ing of  the  column  causes  the  stress  to 
be  distributed  unevenly  over  the  cross 
section  of  the  column  and  makes  it 
greater  on  the  concave  side  than  the 
value  obtained  by  dividing  the  load  by 
the  sectional  area  (see  Fig.  107).  The 
formulas  used  in  designing  columns, 
and  in  calculating  the  stress  developed 
in  them,  are  to  a  large  extent  empirical.  '  I07' 

A  large  number  of  formulas  have  been  developed  by 
different  investigators,  and  those  in  most  common  use 
will  be  given. 

128.  STIFFNESS  OF  COLUMNS.     If  a  flat  board  is  used 
as  a  column,  bending  will  occur  about  an  axis  parallel 
to  the  longer  side  of  a  section.     In  all  columns  free  to 

221 


222  COLUMNS  AND   STRUTS  [CHAP.  XIV 

bend  in  any  direction  bending  will  occur  in  the  direction 
in  which  the  column  is  least  stiff.  In  other  words,  the 
bending  will  occur  about  the  axis  for  which  the  moment 
of  inertia  and  the  radius  of  gyration  are  the  least.  The 
most  economical  column  section,  therefore,  would  be 
one  for  which  the  tendency  to  bend  would  be  the  same 
for  all  axes. 

The  slenderness  ratio  is  the  ratio  of  the  length  of 
a  column  to  the  least  radius  of  gyration  of  the  cross 

section,  and  equals  - »  where  /  is  the  length  of  the  column 

and  r  is  the  least  radius  of  gyration  as  determined  by 
the  principles  of  Appendix  A.  /  and  r  should  be  in  the 
same  units,  and  the  inch  is  the  unit  most  commonly 
employed. 

129.  THE  STRENGTH  OF  COLUMNS.  The  yield  point 
of  the  material,  which  is  somewhat  higher  than  the 
elastic  limit,  is  practically  the  ultimate  strength  for  col- 
umns built  of  structural  steel  or  other  similar  material. 
When  a  column  is  sensibly  bent,  the  bending  moment  at 
the  section  of  greatest  deflection  increases  rapidly  with 
a  small  increase  of  load.  The  moment  of  the  load  at 
the  danger  section  will  cause  the  column  to  fail  under  a 
load  somewhat  greater  than  that  load  which  will  develop 
a  stress  equal  to  the  elastic  limit  of  the  material. 

Fig.  108  shows  characteristic  failures  for  compression 
specimens  of  timber.  The  short  one  shows  oblique 
shear  failure,  the  intermediate  ones  show  failure  in  com- 
pression, and  the  longest  one  shows  failure  due  to  bend- 
ing of  the  column. 

The  condition  of  the  ends  also  has  an  effect  on  the 
strength  of  a  column.  Fig.  109  shows  the  position 
assumed  by  long  homogeneous  columns  under  load,  with 
different  end  conditions:  (a)  with  both  ends  round 


ART.  129]          THE   STRENGTH  OF   COLUMNS  223 


FIG.  108. 


COLUMNS  AND   STRUTS 


[CHAP.  XIV 


or  pivoted,  (b)  with  one  end  round  and  the  other  end 
fixed,  and  (c)  with  both  ends  fixed.  Fixing  the  ends 
increases  the  strength  of  a  column.  In  Fig.  109  it  may  be 
seen  that  about  two-thirds  of  the  column  in  (£)  is  in  a 
condition  similar  to  that  in  (a),  and  that  half  of  the 
column  in  (c)  is  in  a  condition  similar  to  that  in  (a). 


(a)  (b)  (c) 

FIG.  109. 

It  is  commonly  assumed  that  fixing  one  end  is  equivalent 
to  decreasing  the  length  to  f  and  leaving  the  ends 
round,  and  fixing  both  ends  is  equivalent  to  decreasing 
the  length  to  f  and  leaving  both  ends  round. 

In  very  long  columns  the  column  may  fail  by  sidewise 
deflection  without  any  portion  of  the  material  being 
injured.  This  action  occurs  at  a  lower  slenderness 
ratio  in  a  material  like,  timber  than  in  a  material  like 
steel.  The  phenomenon  of  sidewise  failure  can  be 
illustrated  by  the  blade  of  a  tee-square. 


ART.  130]         THE   STRAIGHT-LINE   FORMULA  225 

130.  THE  STRAIGHT-LINE  FORMULA.    By  examining 

the  data  of  tests  of  columns  it  is  found  by  plotting  points 

p 
to  represent  the  average  stress  -7  at  rupture,  for  various 

values  of  the  slenderness  ratio  -»  that  a  straight  line 

can  be  drawn  which  will  fairly  represent  average  ulti- 

p 
mate  values  of  -j  for  the  different  slenderness  ratios. 

A 

p 
Thus,  in  Fig.  no  values  of  -j  are  given  along  the  vertical 

axis  and  values  of  -  are  given  along  the  horizontal  axis. 

An  equation  representing   this  straight  line  is  of  the 

P  C*  I  • 

form  — r  =/i —  •     In  this  formula  PI  is  the  load  at 

A  f 

rupture.  A  similar  formula  may  be  used  to  determine 
the  safe  load  for  a  given  column  or  to  determine  the 
proper  sectional  area 
to  carry  a  given  load. 
Of  course,  the  safe  load 
P  will  be  considerably  -£ 
lower  than  the  rup- 
turing load  PI.  The 
straight-line  formula 
for  the  average  stress  ~ 

over  the  section  of  the  FlG- IIO< 

P             Cl 
column  is  -r  =  / where  P  is  the  safe  load  and  the 

A  T 

values  of  /  and  C  are  to  be  specified.  In  the  straight- 
line  formula /—  C-  is  considered  as  the  allowable  safe 

unit-stress.  This  formula  shows  that  the  strength  of  a 
column  becomes  less  as  the  length  increases  and  as  the 
radius  of  gyration  decreases.  It  is  purely  empirical,  as 


226  COLUMNS   AND    STRUTS  [CHAP.  XIV 

it  is  based  entirely  upon  experimental  data,  but  it  is  con- 
sidered to  be  as  reliable  as  any  and  is  quite  generally 
employed. 

The  values  of  /i  and  C\  for  various  materials  can  be 
derived  from  experimental  data,  and  they  may  be  used 
as  a  guide  to  determine  values  of  /  and  C  to  be  used  in 
design. 

The  greatest  value  of  the  slenderness  ratio  to  be  used 
in  the  application  of  the  straight-line  formula  is  usually 
given  as  I oo  to  125.  In  any  case  it  should  not  be  greater 
than  150. 

P  I 

Cooper  gives  the  formula  -r  =  17,000  —  90-  as  safe 

yl  T 

for   soft   steel    columns   of   a   through   railroad   bridge. 

p  i 

Ketchum  gives  the  formula  -j  =  16,000  —  70-  for  steel 

A.  T 

columns  in  building  frames.     The  limit  of  -  is  125. 

The  Chicago  building  ordinance  as  revised  in  1910 
may  be  taken  to  be  illustrative  of  architectural  practice. 
Its  requirements  are  as  follows : 

P  I 

For  steel  columns -r  =  16,000  —  70  -  • 

J\.  T 

p  I 

For  wrought  iron  columns  .  .  .  -r  =  12,000  —  6o-» 

^T.  T 

p  I 

For  cast  iron  columns -r  =  10,000  —  40  -• 

A  T 

The   maximum   allowable   compressive  stress  shall   not 

exceed  the  values  given  in  Table  18.  -  shall  not  exceed 
120. 

For  timber  columns  the  following  is  a  modification 
of  the  formula  used  by  Ricker  and  of  that  given  in  the 
Chicago  building  ordinance : 

P  I 

For  timber  columns.  . . .  -r  =/  —  .00367 -» 


ART.  130]         THE   STRAIGHT-LINE   FORMULA 


227 


in  which  /  is  the   value   of   the  allowable  compressive 
stress   parallel    to   the   grain   given   in   Table    8.     The 

ordinance  provides  that  the  slenderness  ratio  -  shall  not 

exceed  120.     The  original  form  of  this  formula  was  for 
columns  of  rectangular  section. 

TABLE   1 8 

MAXIMUM  ALLOWABLE   COMPRESSIVE   STRESS 

IN  POUNDS  PER  SQUARE  INCH,  CHICAGO 

BUILDING  ORDINANCE,  1910. 


Material. 

Compressive  stress, 
pounds  per  square  inch. 

Steel 

14,000 

Wrought  iron  
Cast  iron  

IO.OOO 
IO,OOO 

Two  problems  can  be  solved  by  the  use  of  the  straight- 
line  formula, —  (i)  the  safe  load  a  given  column  will 
carry,  and  (2)  the  design  of  columns.  For  these  problems 
the  above  formulas  may  be  used  unless  the  specifications 
state  otherwise.  Of  these  problems  the  design  of  col- 
umns is  the  One  most  commonly  met  by  the  engineer 
and  the  architect,  and  it  admits  of  many  solutions. 

ILLUSTRATIVE   EXAMPLES 

i.  What  load  would  a  1 5-inch,  42-pound  I-beam  9  feet  long 
safely  carry  if  used  as  a  column  in  a  bridge  ? 

From  Table  No.  21  giving  properties  of  I-sections,  A  =  12.48 
square  inches,  the  least  r  =  1.08  inches. 
/  _  QX  12  _  108 
r  ~     i. 08     ~  i. 08  " 
By  the  use  of  Cooper's  formula 

P  =  A  ^17,000  -  90  A 
P  =  12.48  X  (17,000  —  9000)  =  12.48  X  8000  =  99,840  pounds. 


228  COLUMNS  AND   STRUTS  [CHAP.  XIV 

2.   Design  a  square  shortleaf  pine  column  n  feet  long  to  carry 
a  load  of  10,000  pounds. 

Let  d  be  the  dimension  of  a  side  and  r  the  radius  of  gyration. 

A  =  d*. 
The  least  value  of  r  may  be  such  that 

/  12  X  ii       132 

-  =  120,    least  r  =  —     =  -*-  =  i.i  inch. 

r  120  120 


=  v7  +A 

V        -r  n         •        fll 

12 

Least   d  =  r  Vi2  =  i.i  X  3.46  =  3.81  inches. 

P  I 

For  the  formula      —  =  /  —  .00367  - , 

A  T 

/  =  1200  from  Table  8. 
=  1200  -  .0036  X  1200  X  /IX  X  I2> 


10,000  =  i2oodz  —  igjod, 
d2-  i.64</  =  8.33, 
d  —  3.82  inches. 

The  3. 8 2 -inch  by  3. 8 2 -inch  timber  would  do,  but  a  4-inch  by 
4-inch  would  generally  be  used. 

In  this  example  the  size  of  the  column  is  the  same,  determined 
both  by  the  slenderness  ratio  and  by  the  allowable  stress.  This 
is  seldom  the  case. 

131.  ECCENTRIC  LOADS  ON  COLUMNS.  The  fore- 
going formulas  are  to  be  used  only  for  axial  loads.  As 
shown  in  Art.  77,  a  load  when  eccentric  produces  a  greater 
unit-stress  than  when  axial,  and  when  the  load  on  a  col- 
umn is  eccentric  the  formulas  used  must  take  account  of 
the  effect  of  the  eccentricity.  If  the  load  P  has  the  eccen- 
tricity e,  the  stress  due  to  the  eccentricity  alone,  as  de- 

P  ec 

rived  in  Art.  77,  is  -j  -» •     Consequently   the  allowable 
A.  Y 

unit-stress  for  design   of  a   column  must  be  equal   to 


ART.  132]      TRANSMITTING   LOADS  TO   COLUMNS  2 29 

P      P  ec 

—  +  -r  -t ,  and  by  equating  this  to  the  allowable  stress, 

^T.  ./I     T 

the  straight-line  formula  for  columns  carrying  eccentric 
loads  becomes 


The  second  member  of  this  equation  is  the  allowable 
stress  as  given  in  the  formulas  of  Art.  130. 

ILLUSTRATIVE  EXAMPLE 

What  load  would  a  solid  round  cast  iron  column  6  inches  in 
diameter  and  10  feet  long  safely  carry  if  the  load  has  the  eccen- 
tricity of  1.5  inches? 

For  this  column  the  formula  is—fi+         =  10,000  -  40-- 


A  =  —  =  28.3  square  inches,  e  =  1.5  inches,  c  =  3  inches, 
4 

r2  =  I  +  A  =  r--^  —  =  -r=^7  =  2.25,  r  =  1.5  inches. 
64         4        16       16 

/  _   IPX  12  _ 

—  ou, 

r         1.5 

—  1 1  +  T'5  X  3  )  =  10,000  -  40  X  80  «=  6800. 
.3V     2.2.=;  / 


28.3  V~    2.25 
P  =  — —  X  6800  =  64,200  pounds. 

132.  THE  METHODS  OF  TRANSMITTING  LOADS  TO 

COLUMNS.  In  the  columns  of  such  structures  as  bridges 
the  load  is  usually  transmitted  to  the  column  through 
pins  or  rivets  and  plates,  in  such  a  manner  that  the  load 
is  axial  or  so  that  the  eccentric  stress  will  oppose  any 
moment  stress  that  may  be  developed  by  the  weight  of 
the  member  when  not  vertical.  For  buildings  and  many 
other  structures,  however,  the  load  may  be  transmitted 


230 


COLUMNS  AND   STRUTS 


[CHAP.  XIV 


through  angles  and  rivets  on  one,  two,  three,  or  four 
sides  of  the  column,  thus  producing  large  eccentric 
stress  which  should  be  provided  for  in  the  design  of  the 
column.  Beams  and  girders  are  usually  supported  on 
caps  or  brackets,  and  for  purposes  of  design  the  load  is 
considered  as  acting  at  the  centroid  of  the  area  support- 
ing the  member.  The  resultant  of  all  the  loads  may  be 
found  and  dealt  with,  or  the  effect  of  each  load  may  be 
determined  separately  and  the  resulting  stresses  com- 
bined. In  making  the  combination  of  the  loads  it 
should  be  borne  in  mind  that  loads  on  opposite  sides 
of  the  column  will  partly  neutralize  the  eccentric  effect. 


OTHER  COLUMN  FORMULAS 

133.  COMPARATIVE  STRENGTH  AND  STIFFNESS  OF  LONG, 
IDEAL  COLUMNS.  Condition  of  the  Ends.  From  analogy  with 
beams  the  maximum  deflection  for  a  given  stress  in  the  outer 
fiber  is  taken  to  vary  directly  with  the  square  of  the  length 

of  the  column  (Art.  108). 


' 

B 

/ 

I 

\ 

1 

A 

1 

• 

-L             L_ 

B 


(a)          (6)          (c) 


Consequently  the  maxi- 
mum moment  and  the 
maximum  moment  stress 
in  columns  are  assumed 
to  vary  with  the  square 
of  the  length.  For  very 
long,  ideal,  homogeneous 
columns  the  assumptions 
are  approximately  true. 
Columns  under  load  will 
deflect  approximately,  as 
shown  in  the  curves  of 
Fig.  in.  Curve  (a)  is  for 
both  ends  round  or  hinged; 
curve  (b)  is  for  one  end 

round  or  hinged  and  the  other  end  fixed;  curve  (c)  is  for  both 
ends  fixed;  curve  (d)  is  for  one  end  fixed  and  the  other  end  round 


FIG.  in. 


ART.  134]  RANKINE'S  FORMULA  231 

and  free  to  move;  curve  (e)  is  for  both  ends  fixed  in  direction 
but  free  to  move  laterally. 

If /i  is  the  moment  stress  developed  in  a  column  of  length  h  with 
both  ends  round  or  hinged  (represented  by  curve  (a))  under  a  given 
load,  the  moment  stress  developed  in  the  same  column  under  the 
same  load  with  its  ends  fixed  in  the  various  ways  will  now  be  found. 

Fixing  one  end  is  equivalent  to  shortening  the  column  to  f  h 
and  leaving  the  ends  round.  The  portion  AB  of  the  column 
shown  in  curve  (b)  is  in  a  condition  similar  to  a  round-ended  column. 
For  this  case,  one  end  fixed  and  one  end  round,  therefore,  the 
moment  stress  developed  will  be 

/s- 

Fixing  both  ends  is  equivalent  to  shortening  the  column  to  £  h 
and  leaving  both  ends  round.  The  portion  AB  may  be  considered 
as  a  round-ended  column.  The  maximum  moment  stress  devel- 
oped for  this  case  will  be 


py 

W  ,    i, 

-  ~TT^  =  -Ji' 


The  column  with  one  end  fixed  and  the  other  end  round  or  hinged 
is  in  the  same  condition  as  half  of  a  column  with  both  ends  round, 
and  for  this  case  the  moment,  stress  developed  will  be 


By  keeping  both  ends  restrained  in  direction  but  one  end  free 
to  move  laterally,  as  in  (e},  is  equivalent  to  having  two  columns 
similar  to  the  condition  shown  in  (d)  but  one-half  as  long;  there- 
fore the  moment  stress  developed  in  this  case  is 


134.  RANKINE'S  FORMULA.  Columns  of  Intermediate  Length. 
Rankine  derived  an  empirical  formula  for  columns  of  interme- 
diate lengths,  such  as  are  found  most  commonly  in  engineering 


232 


COLUMNS  AND   STRUTS  [CHAP.  XIV 


practice.  The  following  is  the  derivation  (see  Fig.  112):  Let 
the  column  have  the  maximum  deflection  A,  then  the  maxi- 
|p  mum  stress,  which  is  due  to  compression  and 
flexure,  will  be  at  the  point  of  maximum 
deflection  and  is  /  =/x  +/2,  where  fv  is  the 

p 

direct  compressive  stress  which  equals  —  and 

A 

f2  is  the  bending  stress  and  is  — -  (Art.  69). 
Therefore 


,  _  P      Me  _  P 

J      A  +  I    ~A+~T 

From  analogy  with  the  maximum  deflection  of 
beams  (Art.  108)  it  is  assumed  that  A  varies 

I2 
with  - .    If  ^  is  a  factor  depending  upon  the 

material  of  the  column  and  the  condition  of 


the  ends,  A  =  </>-,  then 
c 


Af 


I  =  Ar2,  and  -  is  the  slenderness  ratio.     The  factor  0  is  a  fraction 

TABLE   19 

VALUES  OF  </>    USED   IN   RANKINE'S  FORMULA 


Conditions  of  the  ends. 

Timber. 

Cast  iron. 

Wrought 
iron. 

Steel. 

4 

4 

4 

4 

3000 
I.78 

5000 
1.78 

36,000 
I.78 

25,000 
I.78 

3000 

I 

5000 

I 

36,000 

I 

25,000 
I 

3000 

5000 

36,000 

25,000 

ART.  135]  EULER'S   FORMULA  233 

which  is  determined  partly  by  experiment  and  partly  from  the 
theory  of  Art.  133.  Having  experimental  data  available  and 
assuming  the  relative  strengths  as  given  in  Art.  133  the  values 
of  0  were  found  to  be  as  given  in  Table  19.  It  may  be  noted  that 
trie  numerator  of  the  fraction  indicates  the  condition  of  the  ends 
and  the  denominator  is  the  characteristic  for  the  material.  /  should 
be  the  allowable  working  stress  for  problems  in  design. 

135.  EULER'S  FORMULA.  Long  Columns.  Euler  deduced  a 
formula  for  long,  ideal,  homogeneous  columns.  For  such  columns 
it  has  been  found  that  when  the  load  reaches  a  certain  limit,  if 
a  lateral  deflection  occurs  the  load  will  hold  the  column  in  equi- 
librium in  that  position.  If  the  load  is  decreased  the  column 
will  come  back  to  a  straight  position,  and  if  the  load  is  increased 
the  deflection  increases  until  failure  finally  follows.  From  analogy 
with  beams,  the  deflection  of  a  column  from  the  straight  position 
varies  inversely  with  the  modulus  of  elasticity  and  the  moment 
of  inertia  of  the  section,  and  directly  as  the  square  of  the  length 
(Art.  108);  consequently  the  load  a  given  long  column  will  carry 
is  directly  proportional  to  the  modulus  of  elasticity  and  to  the 
moment  of  inertia  and.  is  inversely  proportional  to  the  square  of 
the  length.  Therefore  the  formula  based  on  these  principles  has 

the  general  form  P  =  ——-  • 

For  n  Euler  deduced  the  theoretical  value  irz  for  columns  with 
both  ends  round,  2|7r2  for  columns  with  one  end  fixed  and  the 
other  end  round,  4?r2  for  columns  with  both  ends  fixed.  Therefore, 
the  values  of  the  load  P  which  will  cause  failure  as  determined  by 
Euler's  formula,  are: 

Both  ends  round, 

ir*EA 


\ri 
One  end  fixed,  the  other  end  round, 


234  COLUMNS'  AND   STRUTS  [CHAP.  XIV 

Both  ends  fixed, 


\r) 
One  end  fixed,  the  other  round  and  free  to  move, 


In  these  formulas  E  is  the  modulus  of  elasticity,  7  =  Ar2  is  the 
least  moment  of  inertia  of  the  cross  section,  r  is  the  least  radius 

of  gyration  of  the  cross  section,  and  -  is  the  slenderness  ratio. 

Euler's  formula  as  given  is  for  the  critical  load,  and  it  should 
be  modified  before  being  used  for  design,  and  it  should  be  used 
only  for  columns  of  which  the  slenderness  ratio  is  not  less  than 
about  200.  For  design  the  formula  would  be  modified  by  intro- 
ducing a  "  factor  of  safety."  (See  Example  4.) 

136.  THE  THREE  PROBLEMS.    Three  typical  problems  may 
be  investigated  by  the  use  of  the  column  formulas:  (i)  The  in- 
vestigation of  columns,  which  consists  of  determining  the  maxi- 
mum unit-stress  developed  in  a  given  column  under  a  given  load. 
(This  can  be  done  only  with  Rankine's  formula,  for  which  case  the 
stress  is  only  nominal.)     (2)  The  load  which  a  given  column  will 
carry  safely.     (3)  The  design  of  a  column  to  carry  a  given  load. 

137.  ECCENTRIC  LOADS  ON  COLUMNS.    Rankine's  and  Euler's 
formulas  as  given  above  are  to  be  used  only  when  the  load  is 
axial.    In  Rankine's  formula, 


p 
for  axial  loads,  the  part  of  the  stress  —  is  due  to  direct  compres- 

jrL 

P        /A2 

sion  and  the  part  —  X  <t>{-)  is  due  to  the  bending  moment  in  the 
A          \rj 

column.    If  the  load  has  the  eccentricity  e,  the  increase  in  the 

stress  due  to  this  eccentricity  by  Art.  77  is  — •  — ;   consequently 

A  Y 


EXAM.]  COLUMNS  AND   STRUTS  235 

the  stress  developed  in  a  column  under  an  eccentric  load  will  be 
the  sum  of  the  three  stresses  and  is 

P   ,  P     /A2  ,   P  ec  P 


which  may  be  used  for  columns  to  which  Rankine's  formula  would 
ordinarily  be  applied  for  an  axial  load. 

For  columns  of  large  slenderness  ratio  a  nearer  approximation 
may  be  made  as  follows:  If  P  is  the  load  with  the  eccentricity  e 
and  the  maximum  deflection  A,  the  total  eccentricity  of  p 

the  load  at  the  point  of  maximum  deflection  is  ei  =  e  +  A 
(Fig.  113),  and  by  considering  the  stress  on  the  cross  sec- 
tion at  that  point  as  the  result  of  the  eccentric  load  P, 
the  maximum  unit-stress  from  Art.  77  is 


in  which  ei  must  be  calculated.  Let  P0  be  the  load  ob- 
tained by  the  use  of  Euler's  formula  for  the  column,  and 
imagine  it  placed  concentric  with  the  column  when  the 
deflection  is  A,  then  the  column  will  be  in  equilibrium 
under  that  load.  As  the  column  is  in  equilibrium  under 
either  the  eccentric  load  P  or  the  concentric  load  P0,  the 
moments  at  the  danger  section  for  both  loads  may  be 
equated,  hence 

P0A=P(*+A)    or,     A--** 


and 


and  /  =  ±-  { i  +  _gP°*L    . ,-  ™   II3. 


This  formula  for  eccentric  loads  on  columns  may  be  used  for 
long  columns  for  which  Euler's  formula  could  be  applied  for 
axial  loads. 


EXAMPLES 

i.   If  two  8-inch,  1 8-pound  I-beams,  latticed  together  so  that 
the  distance  between  their  centroids  is  6^  inches,  are  used  as  a 


236  COLUMNS  AND    STRUTS  [CHAP.  XIV 

column  20  feet  long  with  both  ends  round,  what  is  the  unit-stress 
developed  by  an  axial  load  of  40  tons  ? 
By  Rankine's  formula 

/-5(.+.«m 

A\          VJ I 

P  =  80,000  pounds,  A  =  10.66  square  inches, 

0  =  — — ,  /  =  20  X  12  =  240  inches, 
25,000 


/  _  ^  tl+^-fw}1}  _  7500  x  1.86 

10.66  V        25,000  \3.27/ / 
=  14,000  pounds  per  square  inch. 

2.   If  the  load  in  Example  No.  i  is  applied  2  inches  from  the  cen- 
troid  of  the  section  what  would  be  the  maximum  stress  developed  ? 
For  this  case  the  formula  is 


c  =  (x  -\-  width  of  flange)  -5-  2  =  (6.32  +  4)  -5-  2  =  5.18  inches. 
=  So.ooo/    +      4      /^4Q\2  +  2  X  5.i8\  , 
10.66  V        25,ooo\3.27/         (3-27)*  / 

/  =  7500  (i  -f-  .86  +  .96)  =  7500  X  2.82  =  21,200  pounds  per 
square  inch. 

These  results  show  that  when  a  load  is  axial  the  stress  may  be 
within  the  safe  limit,  while  a  slight  shifting  of  the  load  may  cause 
dangerously  high  stresses. 

3.  Design  a  square  timber  column  10  feet  long  with  one  end 
fixed  and  the  other  end  round  to  carry  a  load  of  5000  pounds  safely. 

Let  d  be  one  dimension  of  the  section,  then  A  =  d2 

I     "d*  d 

r  =  i/—     —=  =  —7—  inches,   /  =  10  X  12  =  120  inches. 
V  12  -r-  d2      via 

For  Rankine's  formula 

<t>  =  ^J—  and  /  =  800  pounds  per  square  inch. 
3000 

5000 800 =       800  d2 

d*    "         1.78  X  120  X  120  X  12      d2  +  102.5 

3000  Xd2 
8d4  —  50  d2  —  5125  =  o. 


ART.  138]      BEHAVIOR  OF   COLUMNS  UNDER  LOAD       237 

Solving  this  as  a  quadratic  equation, 
dz  =  29.4, 
d  =  5.  4  inches. 

For  Euler's  formula,  by  using  a  factor  of  safety  of  10, 


5000  X  io  =  2'25  X  9'87  X  I-5°0'000  X      =  192  d', 

120  X  120  X  12 


d*  =  260, 

d  =  4+  inches. 

As  the  slenderness  ratio  of  the  column  is  in  the  neighborhood 
of  100,  the  result  obtained  by  Euler's  formula  is  not  so  reliable 
as  the  other. 

4.  What  will  be  the  maximum  stress  developed  in  a  cast  iron 
column  4  inches  in  diameter  and  1  2  feet  6  inches  long,  with  both 
ends  round,  carrying  a  load  of  70x30  pounds  placed  i  £  inches  from 
the  center  of  the  end? 

By  the  use  of  Rankine's  modified  formula  for  eccentric  loading, 
we  find  the  stress. 


i  inch, 


/  =  150  inches,  A  =  12.57  square  inches, 

,      7000  /T          4    /i5oA  ,   1.5  X  2\       7000  /T       TQ   .   .\ 

Jf  = i  +  1 )  H 1  =  -    —  (i  +  io  +  3) 

12. 57V         5000  V  i  /  i      /      12.57 

=  12,300  pounds  per  square  inch. 


138.  BEHAVIOR  OF  COLUMNS  UNDER  LOAD.  In 

columns  of  ordinary  length  used  in  construction  the 
stresses  set  up  by  eccentricity  of  loading  due  to  non- 
straightness,  unevenness  of  bearing  at  ends,  and  other 
causes  due  to  shop  and  erection  processes,  often  are  so 
great  that  the  effect  of  the  length  of  the  column  is  almost 
negligible.  This  is  especially  true  of  columns  built  up 


238  COLUMNS  AND   STRUTS  [CHAP.  XIV 

of  several  parts  (e.g.,  a  column  built  up  of  two  channels 
connected  by  lattice  work).  Due  to  bends  in  the  com- 
ponent parts  of  such  built-up  columns,  slip  of  rivets 
and  other  causes,  the  extreme  fiber  stress,  even  in  short 
columns,  may  be  as  much  as  50  per  cent  greater  than  the 
average  stress.*  Furthermore,  in  designing  columns 
great  care  should  be  taken  that  they  are  not  built  up 
of  so  thin  metal  that  there  is  danger  of  failure  by 
"wrinkling"  of  plates  under  load.  So  much  uncer- 
tainty exists  as  to  the  action  of  built-up  columns  that 
low  stresses  should  be  used  in  designing  them,  and  care 
should  be  taken  to  see  that  any  new  column  is  not 
built  up  of  parts  relatively  thinner  and  more  liable 
to  "wrinkling"  failure  than  are  the  parts  of  existing 
successful  columns. 

A  formula  depending  upon  experimental  data  for  its 
constants  should  be  used  only  in  designing  columns 
similar  to  those  from  which  the  data  were  derived.  For 
example,  if  a  series  of  experiments  is  made  upon  columns 
of  one  shape  of  cross  section,  the  data  should  not  be  re- 
lied upon  in  designing  columns  of  a  different  shape  of 
cross  section,  although  the  material  and  slenderness  ratio 
may  be  the  same.  Whether  the  results  of  tests  of  small 
columns  can  be  used  for  determining  the  allowable  stresses 
in  similar  large  columns  is  a  disputed  question  among 
engineers.  Such  a  procedure  is  sometimes  necessary,  and 
in  such  a  case  working  stresses  in  the  large  columns  should 
be  low. 

*  See  Bulletin  No.  44  of  the  Engineering  Experiment  Station  of  the 
University  of  Illinois. 


EXAM.]  COLUMNS  AND   STRUTS  239 

EXAMPLES 

1.  What  should  be  the  distance  from  center  to  center  of  two 
8-inch,  1 8-pound  I-beams  latticed  together  for  a  column  section 
to  make  the  radii  of  gyration  equal  for  the 

two  principal  axes?     (Fig.  114.) 

The  moments  of  inertia  about  the  X  and 
Y  axes  must  be  made  equal.  From  the 
table  of  properties  of  I-sections,  the  mo- 
ment of  inertia  of  one  of  the  sections 
about  the  X-axis  is  56.9  (in.)4,  and  the 
moment  of  inertia  about  the  F'-axisis  3.78 
(in.)4,  and  the  area  of  one  section  is  5.33  pjG 

square  inches.     The  moment  of  inertia  of 

one  section  about  the  F-axis  is  I+Ad  and  equals  3.78  +  5.33  f-.l . 

(fx\*\ 
3.78  +  5.33  f  -  J  j.     For  equal  moments  of 

inertia  of  the  built-up  section  about  the  X  and  F  axes, 

2  (56.9)  =  2  (3-78  +  5-33  (-J, 

.'.  x  =  6.32  inches. 

2.  A  column  built  up  of  two   io-in.,   i5-lb.   channels  laced 
together  with  a  distance  of  6.33  inches  between  the  backs,  is 
1 8  feet  long.     Determine  the  load  the  column  will  carry  with  eccen- 
tricities of  o  inch,  i  inch,  2  inches,  4  inches,  6  inches,  8  inches, 
10  inches,  12  inches,  and  14  inches,  respectively,  the  point  of 
application  of  the  load  being  on  the  centroidal  axis  which  is  per- 
pendicular to  the  web  of  the  channel,  and  plot  the  curve  showing 
the  relation  between  the  load  and  the  eccentricity. 

The  moments  of  inertia  about  both  principal  axes  are  equal 
for  the  given  spacing  and  r  =  3.87  inches,  A  =  8.92  square  inches, 

/  =  12  X  18  =  216  inches,  l=^  =  55.8,  c  =  6'33  +  2  X  2'6 

r       3.87  2 

=  5.77  inches,  £  -  &S'.  .385. 
By  the  use  of  Ketchum's  formula, 

—  ( i  +  —  J  =  16,000  -  70  -  =  16,000  -  70  X  55.8  =  12,100  pounds 
per  square  inch, 


240 


COLUMNS   AND   STRUTS 


[CHAP.  XIV 


P  (i  +  .385  e)  =  8.92  X  12,100  =  107 ,800 pounds. 

.'.  P0  =  107,800  pounds, 
P!  =  107,800  -T-  1.385  =  78,000  pounds, 
P2  =  107,800  -T-  1.77  =  60,900  pounds, 
P4  =  107,800  -j-  2.54  =  42,500  pounds, 
P6  =  107,800  -7-3.31  =  32,600  pounds, 
P8  =  107,800  -T-  4.08  =  26,400  pounds, 
P10  =  107,800  -7-  4.85  =  22,200  pounds, 
P12  =  107,800  -T-  5.62  =  19,200  pounds, 
P14  =  107,800  -7-  6.39  =  16,900  pounds. 

These  results  are  plotted  in  Fig.  115. 


100000 

80000 

1 

360000 
o 
PW 

a 

1340000 
o 
i-l 

20000 

Load  -  eccentricity  Curve 
for  a  column  18  ft.  long 
of  2  10  in.  ®  15  Ib.  channels 
latticed  together. 
K1*!^16000-™^ 

\ 

\ 

t 

| 
\ 

\ 

s 

X 

^v 

^s 

^ 

)  — 

^-t 

>^^ 

"-^<- 

u. 

0  2  i  6  8  10  12  1 

Eccentricity  hi  Inches. 

FIG.  115. 

3.  Design  the  upper  chord  of  a  roof  truss  hi  which  the  maxi- 
mum stress  is  65,100  pounds  compression,  and  the  length  between 
supported  points  is  5  feet.  Use  two  angles  connected  by  f-in. 
gusset  plates  and  f-in.  rivets. 


EXAM.]  COLUMNS  AND   STRUTS  241 

Since  the  member  is  in  compression  the  rivet  holes  need  not 
be  deducted. 

The  least  allowable  r  is  5  X  I2  =  -  inch. 

I2O  2 

By  use  of  a  handbook  we  find  the  properties  of  the  angles. 

For  direct  compression  alone  the  required  area  is  65,100  •*• 
16,000  =  4.07  square  inches.  For  the  column  section  the  area 
must  be  somewhat  greater.  Try  two  5-inch  by  3-inch  by  A-inch 
angles  placed  with  the  short  legs  outstanding.  The  least  r  is 
1.22  inches,  A  =  4.82  square  inches.  The  allowable  stress  is 

—  =  16,000  —  70  X  -  -  =  1  2,  560  pounds  per  square  inch.    The 

/I  1.22 

actual  stress  is  —  =    5>I°°  =  13,500  pounds  per  square  inch. 
A.         4-  20 

This  is  not  safe. 

Try  two  5-inch  by  3^-inch  by  ^-inch  angles. 
The  least  r  is  1.47  inches,  A  =  5.12  square  inches. 
The  allowable  stress  is 

-  =  16,000  -  70  X  —  —  =  13,140  pounds  per  square  inch. 
A  1.47 

The  actual  stress  is 

=     5>IO°  =  12,750  pounds  per  square  inch. 
A.         5-  12 

This  is  safe. 

The  actual  average  unit-stress  is  nearly  equal  to  the  allowable, 
so  use  two  5-inch  by  3^-inch  by  ^-inch  angles  with  short  legs 
outstanding. 

4.  Draw  the  diagrams  representing  the  relation  between  the 
load  and  the  length  of  columns  of  hemlock  for  the  common  rec- 
tangular sections. 

P  I 

By  use  of  the  formula  —  =  /  —  .00367-  the  values  are  obtained. 
A  T 

From  Table  8,  /  =  1000  pounds  per  square  inch.  For  a  2-inch 
by  2-inch  column,  A  =  4  square  inches,  d  =  3.46  r.  The  maxi- 
mum length  for  which  this  section  may  be  used  is  /  =  1  20  r  = 


When  -  =  o,   P  =  4000  pounds.      When  /  =  5   feet  =  60  inches, 


242 


COLUMNS  AND   STRUTS 


[CHAP.  XIV 


Safe  Loads  for  Hemlock 
Columns  by 
£=1000-3.64 


10  15 

Length  of  Column  in  Feet. 
Fig.  116. 


(I000  _  > 


I000  x 


/  =  6o  X  346  =  IQ3 

=  2500  pounds. 

The  other  values  given  in  the  diagram,  Fig.  116,  were  obtained 
in  the  same  manner  as  those  for  a  2  -inch  by  2  -inch  column. 


PROB.] 


COLUMNS  AND   STRUTS 


243 


PROBLEMS 

1.  Determine  the  distance  between  the  backs  of  two  9-in., 
i3.25-lb.  channels  latticed  back  to  back,  for  equal  radii  of  gyration. 

Ans.  5.62  in. 

2.  What  will  be  the  radii  of  gyration  with  respect  to  the  two 
principal  axes  of  a  column  section  built  up  of  two  io-in.,  25-lb. 
I-beams  and  two  f-in.  cover  plates  12  inches  wide? 

3.  In  a  compression  test  of  specimens  of  different  lengths  of 
the  same  piece  of  red  oak  of  cross  section  i|  inches  by  2  inches 
the  following  values  were  obtained : 


Length, 
inches. 

Maximum  load, 
pounds. 

6 

22,000 

12 

18,000 

24 

14,200 

36 

8,000 

Plot  a  curve  showing  the  relation  between  the  average  unit- 
stress  and  the  slenderness  ratio,  and  determine  the  value  of  /i 
and  Ci  at  rupture,  in  the  straight-line  formula. 

Ans.fi  =  8000  Ib.  per  sq.  in. 
Ci  =      63  Ib.  per  sq.  in. 

4.  Design  a  square  longleaf  pine  column  14  feet  long  to  carry 
a  load  of  6  tons. 

5.  Design  a  latticed  column  18  feet  long  built  up  of  two  steel 
channels  to  carry  a  load  of  20  tons. 

6.  What  safe  load  will  a  hollow  cast  iron  column  10  feet  long 
carry  if  the  outside  dimensions  are  6  inches  by  7  inches  and  the 
inside  dimensions  are  4  inches  by  5  inches  ? 

7.  Design  a  steel  column  14  feet  long   to   carry  an  eccen- 
tric load  of  20  tons  applied  2  inches  from  the  outside  of  the 
column. 

8.  What  should  be  the  spacing  of  2 -inch  by  4-inch  timber 
posts  6  feet  long  to  carry  a  platform  on  which  the  maximum  load 
is  to  be  200  pounds  per  square  foot  ? 


244  COLUMNS  AND    STRUTS  [CHAP.  XIV 

9.  Find  the  load  by  Rankine's  formula  that  would  probably 
rupture  a  cast  iron  column  with  fixed  ends,  18  feet  long  and  6  inches 
in  diameter. 

10.  By  the  use  of  Rankine's  modified  formula  for  eccentric 
loads  on  columns  calculate  the  load  that  would  develop  a  unit- 
stress  of  1000  pounds  per  square  inch  in  a  6-inch  by  6-inch  column 

10  feet  long  with  round  ends  for  the  following  eccentricities: 
(a),  o;  (b),  i  inch;  (c),  2  inches;  (d),  4  inches;  and  (e),  6  inches. 
Plot   a  curve  showing  the   relation  between  the  load  and  the 
eccentricity. 

11.  By  use  of  the  straight-line  formula  solve  Problem  No.  10 
if  the  column  is  of  Washington  fir. 

12.  If  a  i2-in.,  4o-lb.  I-beam  18  feet  long  is  used  as  a  column 
with  round  ends,  what  is  the  slenderness  ratio?    According  to 
Euler's  formula,  what  load  would  cause  rupture  ? 

13.  What  safe  load  will  a  column  27  feet  long  built  up  of  two 
g-in.,  13.25-^.   channels  latticed  together  and  placed  6  inches 
back  to  back,  safely  carry  if  used  in  a  bridge?         Ans.  67,300  Ib. 

14.  What  should  be   the  greatest  length  for  which  timber 
columns  of  the  following  sections  may  be  used?     2  inches  by 
2  inches,  4  inches  by  4  inches,  4  inches  by  6  inches,  6  inches  by 
6  inches,  6  inches  by  10  inches,  6  inches  by  12  inches,  8  inches 
by  8  inches,  10  inches  by  12  inches,  12  inches  by  12  inches. 

Ans.  I  =  34.7  where  d  is  the  least  lateral  dimension.    4  in.  X 
4  in.,  n  ft.,  7  in. 54  in. X  6  in.,  n  ft.,  7  in.;  10  in.  X  12  in.,  28  ft, 

11  in. 

15.  Determine  the  safe  load  for  4-inch  by  4-inch  red  oak 
columns,  which  are  3   feet,  7  feet,  and  n  feet,  7  inches  long 
respectively.    Plot  a  curve  showing  the  relation  between  the  load 
and  the  length  of  the  column.     Do  the  same  for  various  other 
sections  of  oak  columns  carrying  the  curves  to  the  maximum 
allowable  length  of  column.     (This  set  of  curves  may  be  made 
to  include  all  commercial  sizes  of  sections  and  put  on  one  diagram. 
Then  the  red  oak  column  necessary  for  any  load  and  any  length 
can  be  selected  directly  from  the  diagram.) 

16.  Determine  the  safe  load  for  various  lengths  and  various 
sections  of  columns  of  the  different  kinds  of  timber  given  in 
Table  8,  and  plot  the  curves  as  in  Problem  No.  15. 


PROB.]  COLUMNS  AND   STRUTS  245 

17.  Design  a  strut  12  feet,  9  inches  long  in  a  roof  truss  to  carry 
a  compression  load  of  12,000  pounds.     Use  two  angles  with  a  f-in. 
gusset  plate  between  them,  and  f-in.  rivets. 

Ans.   4  in.  X  3  in.  X  TS  in.  [s_. 

1 8.  Design  a  strut  5  feet,  9  inches  long  in  a  roof  truss  to  carry 
a  load  of  20,800  pounds. 

Ans.  Two    2  \  in.  X  2  in.  X  T\in-  ls_,  short  legs  outstanding, 
f-in.  gusset  plate. 

19.  What  four  angles  with  the  long  legs  outstanding  would  be 
required  to  be  riveted  to  a  iVin.  plate  for  a  column  18  feet  long  to 
carry  a  load  of  27,370  pounds? 

Ans.   4  in.  X  3  in.  X  T'S  in.  [s_,  width  of  plate  8  in. 

20.  A  wooden  stick  3 -inch  by  4-inch  in  cross  section  and  10 
feet  long  is  used  as  a  column  with  fixed  ends.     Find  by  Rankine's 
formula  the  unit-stress  developed  under  a  load  of  |  ton. 

21.  Find  the  safe  load  for  a  hollow  cast  iron  column  of  outside 
dimensions  8  inches  by  6  inches,  inside  dimensions  6  inches  by 
4  inches  and  12  feet  long. 

22.  A  hollow  yellow  pine  column  of  square  section,  5  inches 
outside  dimensions,  and  4  inches  inside  dimensions,  has  a  length 
of  1 6  feet.    What  load  could  the  column  safely  carry? 

23.  A  cylindrical  steel  column  with  round  ends  is  36  feet  long 
and  6  inches  in  diameter.     Calculate  by  Euler's  formula  the  axial 
load  that  would  probably  produce  rupture. 

24.  Determine   the  safe  load   for  a  hollow  round   cast  iron 
column  of  external  diameter  1 2  inches,  thickness  i  inch,  and  length 
12  feet. 

25.  A  square  white  oak  column  12  feet  long  is  to  support  a 
load  of  16  tons.    What  must  be  the  size  of  the  column? 

26.  Determine  the  size  of  a  rectangular  loblolly  column  20  feet 
long  to  carry  safely  a  load  of  24  tons.          Ans.   8  in.  by  10  in. 

27.  A  round  solid  cast  iron  strut  15  feet  long  carries  a  load  of 
10  tons.    What  should  be  its  diameter? 


CHAPTER   XV 
TORSION 

139.  STRESS  AND  DEFORMATION.   ROUND  SHAFTS. 
When  a  couple,  as  indicated  by  Pa  in  Fig.   117,  in  a 


FIG.  117. 

plane  perpendicular  to  the  axis  of  a  shaft  acts  upon  the 
shaft,  it  is  twisted,  and  one  cross  section  tends  to  slip 
by  the  section  next  to  it.  This  tendency  is  resisted  by 
the  torsional  stresses  set  up  in  the  shaft.  The  stresses 
developed  are  shearing  stresses.  If  AB  in  Fig.  115  is 
the  original  position  and  AB'  the  final  position  of  an 
element  of  the  surface  of  the  shaft,  the  end  of  the  shaft 
has  twisted  through  the  angle  <£  or  BOB' ,  which  is  pro- 
portional to  the  couple  acting  on  the  shaft  and  to  the 
length  of  the  shaft,  when  the  stresses  developed  are 
within  the  elastic  limit.  The  element  will  have  twisted 
through  the  angle  6  or  BAB',  which  is  proportional 
to  the  couple  but  independent  of  the  length  of  the 
shaft. 

246 


ART.  140]  THE  TORSION  FORMULA  247 

140.  THE  TORSION  FORMULA.   ROUND  SHAFTS.    In 

Fig.  118  let  the  forces  producing  the  couple  be  P  and  the 
arm  between  them  be  p,  then  the  couple  C  equals  Pp. 
Under  the  influence  of  this  couple  the  radius  OA  will 
have  swept  through  the  angle  AOA'  to  the  position  OA ' 
while  it  still  remains  a  straight  line.  The  deformation 


FIG.  1  1  8. 

of  a  fiber  of  the  section  is  proportional  to  the  distance 
from  the  center  0  of  the  shaft  to  the  fiber.  Therefore 
the  unit-stress  developed  on  the  fibers  when  the  greatest 
stress  is  below  the  elastic  limit  is  proportional  to  the  dis- 
tance of  the  fiber  from  the  center.  If  s  is  the  maximum 
unit-stress  developed  upon  the  outer  fiber  of  the  shaft, 
and  r  is  the  radius  of  the  shaft,  the  unit-stress  on  the 

A/ 

fiber  a  distance  y  from  the  axis  is  sy  =  s  -  •     The  total 

§ 
stress  on  the  elementary  area  a  is  -  ya,  and  the  moment 

of  this  stress  about  the  axis  of  the  shaft  is  -  yay  =  -  ay2. 

r  r 

The  moment  of  the  stresses  acting  on  the  entire  cross 
section  is  the  sum  of  all  such  expressions,  and  for  equi- 
librium, 


sJ  Cr 

C    •-  —  ,    or     5  =  -j  , 


248  TORSION  [CHAP.  XV 

where  C  =  Pp  and  is  the  twisting  moment,  and  2  ayz  =  J 
and  is  the  polar  moment  of  inertia  of  the  section  about 
the  axis. 

For  solid  circular  shafts,  Fig.  119, 

r       ird*       Trf4  ,  A  .  .      A  x 

J  =  —  =  —  (Appendix  A), 


c  — 


2r         2          16 
i6C 


FIG.  119.  FIG.  1  20. 

For  hollow  circular  shafts  of  outer  radius  r  and  inner 
radius  ri,  Fig.  120, 


STT  -   ri  57T  - 


s 


2r  16  d 

i6Cd 


7T  - 


Three  typical  problems  may  be  investigated  by  the 
use  of  the  torsion  formula:  (i)  The  investigation, 
(2)  determining  the  allowable  couple,  and  (3)  the  design 
of  a  shaft  to  transmit  a  given  couple. 

141.  STIFFNESS  OF  SHAFTS.  The  relation  between 
the  angle  of  twist  and  the  shearing  modulus  of  elasticity 


ART.  142]  CROSS   SECTION   OF   SHAFTS  249 

may  now  be  deduced.  Since  BB '  in  Fig.  121  is  small, 
BB'  =  r$  =  Id,  <f>  and  6  being  in  radians.  The  detru- 
sion  of  a  fiber  on  the  surface  in  the  length  I  is  BB' ',  the 


FIG.  121. 

D  T3t  i 

unit  detrusion  is  —j—  =  -y-  •     The  shearing  modulus  of 

elasticity  is 

Cr 
Unit-stress  s       J        Cl 


E. 


Unit  detrusion       e       r$ 

an 
ut 


In  the  formulas,  Ea  is  in  pounds  per  square  inch,  C  is 
in  pound-inches,  /  is  in  inches,  /is  in  (inches)4,  and  <j>  is  in 
radians.  In  tests,  if  the  angle  of  twist  is  measured 
in  degrees  the  value  must  be  reduced  to  radians  by  the 
relation 

One  radian  =  57.3  degrees. 

142.  OTHER  SHAPES  OF  CROSS  SECTION  OF  SHAFTS. 

For  any  other  than  circular  sections  the  foregoing 
formulas  cannot  be  applied.  Experiment  has  shown 
that  if  the  section  has  two  axes  of  symmetry  the  fibers 


250 


TORSION 


[CHAP.  XV 


at  the  ends  of  the  shorter  axis  have  the  greatest  dis- 
tortion, and  consequently  the  greatest  unit-stress  will 
occur  at  those  points.  Along  the  cor- 
ners of  rectangular  shafts  there  is  no 
relative  distortion  of  the  fibers,  and 
those  fibers  have  no  stress  developed  in 
them.  Saint  Venant  has  investigated 
the  subject  and  devised  formulas  which 
reduce  to  the  following  forms  for  the 
maximum  stress,  where  C  is  the  twisting 
moment  and  s  is  the  maximum  stress: 
(a)  Square  shaft,*  C  —  0.208  d3s.  s  is  at 
the  middle  of  the  side. 


(a) 


C  =  1-  atfs. 
16 


(b)  Elliptical   shaft,* 

at  the  end  of  the  shorter  axis. 

(c)  Rectangular  shaf  t ,  *  C  =  ( 


s  s 


i.Sb 


•s  is  at  the  middle  of  the  longer  side. 
Merrimanf   gives  for    rectangular  cross    sections    the 
formula  C  =  f  ab2s. 

143.  POWER  TRANSMITTED  BY  SHAFTS.  The  pri- 
mary purpose  of  shafting  is  the  transmission  of  power. 
The  pulleys  are  frequently  fastened  to  the  shaft  by  keys 
and  keyways,  in  which  case  the  formula  for  the  relation 
between  the  maximum  stress  and  the  twisting  moment 
is  complex.  However,  the  power  a  circular  shaft  with- 
out a  keyway  will  transmit  can  easily  be  obtained  if 
the  allowable  stress  is  known.  If  C  is  the  couple  acting 
on  the  shaft  the  work  done  by  turning  the  shaft  through 
an  angle  B  is  CO.  Proof:  Let  P  be  the  force  of  the  couple 

*  See  "History  of  Elasticity,"  Vol.  II,  parti,  by  Todhunter  and 
Pearson. 

t  See  Merriman's  "Mechanics  of  Materials." 


EXAM.]  TORSION  251 

and  p  the  arm.  The  distance  through  which  P  will 
move  in  turning  through  an  angle  0  is  pd,  and  the  work 
done  is  PpB,  or  C6,  as  Pp  =  C.  If  the  shaft  makes  N 
revolutions  per  minute  the  work  done  in  one  minute 
will  be  C2wN. 

C2irN  sJ2irN    =     sJN 

~  33,000  X  12  ~  396>°oo  r  ~  63,030  r  ' 

144.  COMBINED  TWISTING  AND  BENDING.  If  a  bend- 
ing moment  is  developed  in  the  shaft  as  well  as  a  twist- 
ing moment,  there  is  a  combination  of  stresses.  The 
maximum  fiber  stress  developed  by  the  bending  moment 
may  be  obtained  by  the  use  of  the  moment  formula 

/  =  — ,  and  the  maximum  shearing  stress  may  be  ob- 
tained from  the  torsion  formula  s  =  -=- .  These  stresses 

may  be  combined  to  obtain  the  maximum  shearing 
stress  and  the  maximum  tensile  or  compressive  stress 
by  the  formulas 

Sr,   = 


EXAMPLES 

i.   A  solid  circular  steel  shaft  10  feet  long  and  2  inches  in 
diameter  has  a  couple  of  126,000  pound-inches  acting  upon  it. 

(a)  What  is  the  maximum  unit-stress  developed  in  the  shaft? 

(b)  What  is  the  unit-stress  f-inch  from  the  axis  of  the  shaft? 

(c)  At  300  R.P.M.  what  is  the  horse  power  developed  ?     (d)  What 
is  the  angle  through  which  one  end  would  twist  past  the  other? 
(e)  Through  what  angle  would  a  line  on  the  surface  twist? 


252  TORSION  [CHAP.  XV 


(a)        ,  =          =  l6  X>6°°  =  8000  Ib.  per  sq.  in. 


(b)  sv  =  X      =  6000  Ib.  per  sq.  in. 

i          4 

(c)  H.P.  =  ",6°QXarX3oo  .  go.  H.P. 

33,000  X  12 


(e)         0  =  i  X(4°35')  -120  =  2^'. 

2.  What  should  be  the  diameter  of  a  solid  shaft  to  transmit 
500  horse  power  at  80  revolutions  per  minute  if  the  maximum 
torsional  stress  is  not  to  exceed  9000  pounds  per  square  inch? 

sJN 


H.P.  = 

500  = 


63,030  r 

9000  Trtf    X  80 


63,030  X  1 6 
d  =  6  inches. 


PROBLEMS 

1.  What  maximum  unit-stress  will  be  developed  in  a  hollow 
shaft  of  3  inches  outside  and  2  inches  inside  diameter  when  twisted 
by  a  force  of  3000  pounds  at  a  distance  of  i  foot  from  the  axis? 
What  is  the  minimum  stress  developed  ? 

Ans.   8460  Ib.  per  sq.  in. 

2.  What  horse  power  will  be   transmitted  by  the  shaft  in 
Problem  No.  i  when  making  90  revolutions  per  minute? 

3.  What  must  be  the  diameter  of  a  solid  steel  shaft  to  transmit 
1 20  horse  power  at  80  revolutions  per  minute  if  the  allowable  unit- 
stress  is  10,000  pounds  per  square  inch.  Ans.  3.6  in. 


PROS.]  TORSION  253 

4.  If  the  shaft  of  Problem  No.  2  is  20  feet  long  between  the 
pulleys,  what  will  be  the  angle  of  twist  when  transmitting  the  re- 
quired power? 

5.  A  wrought  iron  shaft  7  feet,  6  inches  long  and  2  inches  in 
diameter  twists  through  an  angle  of  10°  30'  under  the  influence 
of  a  couple  produced  by  a  force  of  2500  pounds  at  a  distance 
of   i   foot  from  the  axis.     Compute   the   shearing  modulus  of 
elasticity. 

6.  What  are  the  maximum  shearing  and  tensile  stresses  de- 
veloped in  a  shaft  2\  inches  in  diameter  under  a  twisting  moment 
of  12,000  pound-inches  and  at  the  same  time  under  a  bending 
moment  of  800  pound-feet  ? 

7.  What  will  be  the  maximum  stress  developed  in  a  rectangular 
shaft  of  dimensions  i  inch  by  i  \  inches  if  the  twisting  moment  is 
400  pound-feet? 

8.  Determine  the  maximum  stress  developed  in  a  shaft  i  inch 
square  if  the  twisting  moment  is  produced  by  a  force  of  75  pounds 
at  a  distance  of  14  inches  from  the  axis. 

9.  What  stress  will  be  developed  in  an  elliptical  shaft  of  dimen- 
sions i  inch  by  i \  inches  if  the  twisting  couple  is  400  pound-feet? 

10.  What  should  be  the  diameter  of  a  steel  shaft  to  transmit 
safely  500  horse  power  at  150  revolutions  per  minute? 

11.  Calculate  the  horse   power  that  a  round,  wrought  iron 
shaft  8  inches  in  diameter  and  making  150  revolutions  per  minute 
will  safely  transmit. 

12.  A  hollow  steel  shaft  of  outside  diameter  6  inches  safely 
transmits  450  horse  power  at  100  revolutions  per  minute.     Find 
the  inside  diameter.  Ans.  di  =  3.82  in. 

13.  Find  the  shearing  modulus  of  elasticity  of  a  cast  iron  bar 
10  inches  long  and  0.82  inch  in  diameter  if  twisted  through  an 
angle  of  1.3°  by  a  twisting  moment  of  50  pound-feet. 

14.  A  structural  steel  shaft  120  feet  long  and  16  inches  in  diam- 
eter transmits  8000  horse  power  at  20   revolutions  per  minute. 
Find  the  angle  of  twist  and  the  stress  developed. 

15.  A  solid  shaft  6  inches  in  diameter  is  coupled  by  bolts  i  inch 
in  diameter  on  a  flange  coupling.     The  centers  of  the  bolts  are 
5  inches  from  the  axis.     Find  the  required  number  of  bolts. 

16.  A  wrought  iron  shaft  is  subjected  simultaneously  to  a 


254  TORSION  [CHAP.  XV 

bending  moment  of  10,000  pound-inches  and  a  twisting  moment 
of  12,000  pound-inches.  Determine  the  least  diameter  of  the 
shaft  if  the  maximum  tensile  stress  is  not  to  exceed  10,000  pounds 
per  square  inch  and  the  shearing  stress  is  not  to  exceed  8000 
pounds  per  square  inch. 

17.  Find  the  horse  power  that  can  be  transmitted  safely  by  a. 
cast  iron  shaft  3  inches  in  diameter  and  making  60  revolutions  per 
minute. 

18.  A  steel  wire  0.18  inch  in  diameter  and  10  inches  long  is 
twisted  through  an  angle  of  9.2°  by  a  moment  of  20  pound-inches. 
Determine  the  shearing  modulus  of  elasticity  of  the  wire. 


CHAPTER  XVI 
REPEATED  STRESSES,  RESILIENCE,  HYSTERESIS,  IMPACT 

145.  REPEATED  STRESSES.  The  behavior  of  mate- 
rials under  repeated  stresses  and  impact  is  somewhat 
different  from  that  for  static  or  slowly  applied  stresses. 
The  experiments  of  Wohler,  Bauschinger,  and  others 
for  repeated  stresses  show  that  a  material  will  fail 

TABLE   20 
TESTS  ON  WROUGHT  IRON 

[Wohler.] 


Number  of 
applications. 

Unit-stress  producing 
rupture. 

800 
107,000 

52,800 
48,400 

450,000 

39,000 

10,140,000 

35.000 

under  stresses  lower  than  the  ultimate  strength  of  the 
material.  For  an  enormous  number  of  applications  of 
a  stress  about  equal  to  the  elastic  limit,  the  material 
ruptured.  When  the  stress  was  reversed  and  carried  to 
about  one-half  to  two-thirds  the  elastic  limit  for  each 
reversal,  an  enormous  number  of  applications  of  the 
stress  caused  rupture.  These  experiments  were  carried 
on  in  such  a  manner  that  the  time  between  each  appli- 
cation or  reversal  of  stress  was  so  short  that  the  specimen 
had  no  time  to  rest.  It  is  interesting  to  note  in  Table  20 
the  variation  in  the  maximum  applied  stress  with  the 

255 


256 


REPEATED   STRESSES,   ETC.         [CHAP.  XVI 


number  of  applications  for  wrought  iron.  Fig.  123 
shows  graphically  the  number  of  applications  of  a  given 
stress  necessary  to  produce  rupture  in  wrought  iron. 


100000   200000  300000  400000    500000 
Number  of  Applications  of  Stress. 

FIG.  123. 

Instances  in  which  repeated  stress  and  reversed  stresses 
would  influence  the  design  of  the  members  would  be 
shafting,  car  axles,  piston  rods,  all  rolling  or  vibrating 
members,  etc. 

146.  RESILIENCE.  When  a  load  is  applied  to  a 
member  it  will  deform.  On  removing  the  load  the  mem- 
ber will  resume  its  former  size  and  shape  for  stresses 
below  the  elastic  limit.  And  when  the  elastic  limit  has 
been  exceeded  the  material  will  partly  recover  its  original 
size  and  shape.  The  load  applied  to  the  material  does 
work  on  it,  and  in  turn  when  the  load  is  being  released 
the  material  gives  out  energy.  Resilience  is  the  amount 
of  potential  energy  stored  in  a  material  when  it  is  under 
stress.  Elastic  resilience  is  the  amount  of  potential 
energy  stored  in  a  member  when  the  stress  is  within 
the  elastic  limit.  The  modulus  of  resilience  is  the 
amount  of  energy  stored  in  a  unit  of  volume  of  a  member 
when  the  stress  is  at  the  elastic  limit.  Resilience  can  be 
recovered  to  do  work. 


ART.  146] 


RESILIENCE 


257 


When  the  stress  is  carried  beyond  the  elastic  limit 
permanent  set  is  developed.  In  such  cases  a  larger 
amount  of  work  has  been  done  upon  the  specimen  than 
it  will  give  out  upon  releasing  the  load.  The  work  that 
cannot  be  recovered  is  used  in  permanently  distorting 
the  material,  and  is  converted  into  heat.  Fig.  124  shows 


G  H  F 

FIG.  124. 

a  typical  soft  steel  stress-deformation  diagram.  In  this 
diagram  the  ordinates  represent  the  unit-stress  and  the 
abscissas  represent  the  unit  deformation .  The  work  done 
on  the  material  is  the  average  force  times  the  distance 
through  which  the  force  acts.  Since  the  stress-deforma- 
tion diagram  shows  the  unit-stress  developed  in  a  speci- 
men and  the  corresponding  unit  deformation  under  that 
stress,  the  area  between  the  curve  and  the  horizontal 
axis  represents  the  work  done  on  a  unit  of  volume  of 
the  material.  When  the  stress  is  not  carried  beyond 
the  elastic  limit  all  the  work  done  can  be  recovered. 
The  triangular  area  ACB  represents  the  modulus  of 
resilience.  When  the  point  D  is  reached  the  work  done 
on  a  unit  of  volume  of  the  material  is  represented  by  the 
area  ACDH,  and  the  work  that  can  be  recovered  (the 
resilience)  for  that  point  is  represented  by  the  area 
GJDH.  When  the  point  of  rupture  E  is  reached,  the 


258  REPEATED   STRESSES,  ETC.         [CHAP.  XVI 

total  work  done  on  a  unit  of  volume  of  the  material  is 
represented  by  the  area  A  CDEF. 

147.  RESILIENCE  OF  A  BAR  UNDER  DIRECT  STRESS. 

For  tension  or  compression,  let  the  load  on  the  bar  be  P, 

the  sectional  area  A,  the  length  /,  the  deformation  e. 

For  stresses  below  the  elastic  limit  the  work  done  is 

Pe  _fAfl_P_A1 

2    "    2E         2E 

This  work  done  on  the  specimen  equals  the  resilience 
stored  in  the  specimen.     Therefore  the  resilience  is 


f2 

The  resilience  per  unit  of  volume  is  ^~-    If  /  is  equal  to 

the  elastic  limit  the  resilience  per  unit  of  volume  is  the 
modulus  of  resilience. 

148.  RESILIENCE  OF  A  BEAM.  An  expression  for  the 
resilience  of  a  beam  may  be  deduced  similarly  to  the 
following  method.  Take  the  case  of  a  cantilever  beam 
of  length  /  with  a  concentrated  load  W  at  the  end, 

W 
Fig.  125.     The  average  force  will  be  —  and  the  deflec- 


FIG.  125. 

tion  will  be  A.     The  work  done  on  the  beam  which  is 
equal  to  the  resilience  is 

W        W 


wv 


ART.  149]  MECHANICAL  HYSTERESIS  259 

This  can  be  expressed  in  terms  of  the  maximum  stress 
on  the  outer  fiber  from  the  formula 

,      Me      Wlc  ,J7      fl 

/-—-—•    or    fF-jg. 


fz 
-~  is  the  same  expression  as  obtained  in  Art.  147. 

In  the  case  of  a  uniform  load  each  elementary  load 
does  an  amount  of  work  equal  to  one-half  the  load  times 

the  distance  it  deflects,  and  is  —  -  y  where  w  is  the  load 

per  unit  of  length,  u  is  an  element  of  length,  and  y  is 

the  deflection  at  the  point  (see  Fig.  126).     As  the  load 

is  uniform  the  work  done 

by  each  element  is  pro-  0| 

portional  to  its  deflection 

y.     From  Fig.   126  it  is 

seen  that  uy  is  a  small  FlG  I26 

area  between  the  X-axis 

and  the  elastic  curve  of  the  bent  beam.     The  total  work 

done  is  the  summation  of  all  such  expressions  as  —  y 
and  equals 

^  wu        w^  w  ^  ,  . 

7,  —  y  —  —  2^uy  =  —  X  (area  between  the  X-axis 

^^    22  2 

and  the  elastic  curve) 

since  2  uy  is  the  area  between  the  X-axis  and  the  elastic 
curve.  This  area  may  be  determined  by  the  same  method 
as  is  used  in  finding  the  deflection  curves. 

149.  MECHANICAL  HYSTERESIS.  In  Fig.  127  is  shown 
the  stress-deformation  curve  for  the  case  where  the 
elastic  limit  has  been  exceeded.  After  the  point  A  had 
been  reached  the  load  was  removed.  The  curve  is  convex 
downward,  as  ADC  indicates.  On  reapplying  the  load  the 


260 


REPEATED   STRESSES,   ETC.          [CHAP.  XVI 


curve  will  be  convex  upward,  as  CEA.     The  resilience 

or  work  obtained  from  the 
material  is  A B CD,  and  that 
put  into  it  is  ABCE.  The 
energy  represented  by  the 
loop  ADCE  is  lost  as  heat 
and  is  called  mechanical 
hysteresis. 

150.   LAG.      For  some 
~"  materials   at    stresses    be- 
yond the  elastic  limit  when 


c 
FIG.  127. 


the  load  is  stopped  the  specimen  will  continue  to  deform 
for  some  time.  The  metal 
yields  while  the  load  is 
not  increased.  This  phe- 
nomenon is  known  as  lag, 
and  Fig.  128  is  a  stress- 
deformation  diagram  in 
which  lag  is  shown. 

151.  THE  EFFECT  OF 

REST.     By  allowing  a  FIG.  128. 

specimen  to  rest  after  being  stressed  beyond  the  elastic 

limit,  it  will  partly  re- 
cover its  elastic  prop- 
erties. Fig.  129 
shows  the  stress-def- 
ormation curves  for 
steel  before  and  after 
resting.  In  the  one 
marked  "before  rest- 
ing" the  stress  had 
been  carried  beyond 
the  elastic  limit,  and  reversed  several  times,  the  speci- 
men being  heated  by  the  work  done  on  it. 


FIG.  129. 


ART.  152] 


SUDDENLY   APPLIED    LOADS 


261 


152.  SUDDENLY  APPLIED  LOADS.    In  the  foregoing 

portion  of  the  book  the  load  was  considered  to  be  grad- 
ually applied  to  the  specimen  or  member.  If  the  load 
is  suddenly  applied  the  stresses  are  much  higher  than 
when  the  load  is  gradually  applied.  In  order  to  get  the 
relation  between  the  stress  produced  by  a  gradually 
applied  and  a  suddenly  applied  load  let  the  deforma- 
tion under  the  load  gradually  applied  be  e,  and  when 
suddenly  applied  be  e\.  Having  the  deformation,  the 
corresponding  unit-stress  developed  can  be  determined, 
since  the  stress  below  the  elastic  limit  is  proportional  to 
the  deformation.  The  work  done  on  the  member  when 
the  load  is  gradually  applied  is  equal  to  the  product  of 

We 
the  average  force  and  the  deformation  and  is  - —  .    The 

force  varies  from  o  to  W.     The  area  OA B  in  Fig.  130 


w 


represents  the  work  done.  When  the  load  is  suddenly 
applied  the  total  load  acts  through  the  entire  deforma- 
tion, as  indicated  by  the  line  AB  in  Fig.  131,  but  the 
internal  resisting  stresses  vary  from  zero  to  the  value 
of  FB  along  the  line  OB.  When  the  point  B  is  reached 
the  external  work  done  is  We,  while  the  work  stored  in 

We 
the  member  or  the  resilience  is  - — .     According  to  the 

principle  of  the  conservation  of  energy  the  load  will  not 
stop  until  the  resilience  equals  the  work  done,  conse- 
quently the  deformation  and  the  stress  in  the  member 


262 


REPEATED   STRESSES,   ETC.         [CHAP.  XVT 


will  still  increase.     When  the  deformation  is  e  the  work 
done  in  excess  to  the  resilience  stored  in  the  material  is 

represented  by  the  area  OAB  which  equals—^.    There- 
fore the  excess  of  resilience  over  the  work  done  beyond 


w 


FIG.  131. 

the  point  B  which  is  represented  by  the  area  BCD,  must 

We 
be  equal  to  —  .     The  external  work  done  is  represented 

by  the  area  A  CEO  and  the  resilience  stored  in  the  mate- 
rial is  represented  by  the  area  ODE.  As  the  two  triangles 
BOA  and  BDC  are  similar  and  equal,  the  similar  sides 
must  be  equal,  therefore, 

BC  =  AB  =  e, 


and  consequently  f\  =  2  f  where  /i  is  the  stress  due  to 
the  suddenly  applied  load  and  /  is  the  stress  developed 
when  the  load  is  gradually  applied.  This  shows  that 
the  deformation  and  the  stress  developed  by  a  load  when 
suddenly  applied  are  twice  what  they  would  be  if  the 
load  is  gradually  applied. 

153.   IMPACT  LOADS.     A  load  W  moving  horizontally 

Wv2 
with  a  velocity  v  possesses  the  kinetic  energy  -  —  ,  which 


ART.  153]  IMPACT  LOADS  263 

is  equal  to  Wh,  where  g  is  the  acceleration  due  to  gravity 
and  h  is  the  vertical  distance  the  weight  would  fall  to 
acquire  the  velocity  v.  This  energy  must  be  overcome 
by  the  resilience  stored  in  the  member.  This  energy  may 
be  equated  to  the  resilience  of  the  member  for  any  given 
case  to  obtain  the  stress  developed.  For  example,  if  the 
member  is  in  direct  tension  or  compression 


,      4  1 2  WhE 

f==\-^TJ 

fl 

€l       E' 

A  method  more  generally  applied  is  to  obtain  the 
relation  between  the  deformation  the  load  would  pro- 
duce when  gradually  applied  and  the  deformation  pro- 
duced under  impact.  The  stress  is  proportional  to  the 
deformation.  Let  Q  be  the  maximum  total  resisting  force 
under  the  impact  load  and  e\  the  deformation  produced 
by  the  impact  load.  The  work  done  by  the  resisting 

force  is  — L  since  the  resisting  force  varies  directly  from 

zero  to  Q.     This  work  is  the  resilience  and  equals  the 
external  work, 

2 

If  the  deformation  under  the  static  load   W  is  e  the 
following  proportion  results: 

e       e\ 
Solving  these  two  equations  for  Q  and  e\, 

e\=  A/2  he, 

(3  =  1 


264  REPEATED   STRESSES,   ETC.         [CHAP.  XVI 

From  these  equations  it  is  seen  that  the  deformation, 
and  the  resisting  force,  and  the  unit-stress  developed, 
increase  directly  with  the  velocity  of  the  load,  or  with 
the  square  root  of  the  height  h. 

154.  DROP  LOADS.  If  the  impact  load  falls  verti- 
cally onto  a  member  through  the  height  h  before  im- 
pinging upon  it,  the  load  also  does  work  through  the 
deformation  of  the  member.  Then,  using  the  same 
nomenclature  as  given  in  the  previous  article  we  have, 


Q      W 

and  -  =  —  • 

d       e 

Solving  these  equations  for  e\  and  Q, 

e\  =  e  +  \/2  he  +  e2, 

n      TJ/  ,   W  A/2  he  +  e2 
and  Q  =  W  H  --  -  • 

c 

It  is  seen  from  these  two  equations  that  a  drop  of  a  short 
distance  develops  a  high  stress  compared  with  that 
developed  under  the  static  load  W. 


EXAMPLE. 

i.  Find  the  amount  of  work  necessary  to  stress  a  bar  of 
wrought  iron  5  feet  long  and  i  inch  in  diameter,  from  zero  to 
the  elastic  limit  100  times. 

P  =  25,000  X  .7854  =  19,635  pounds. 


25,OOO,OOO 


Work  =  -  PeN  =  1^35  x  >005  x  I00  =  4009  ft>.lb> 

2  2 


PROB.]  REPEATED   STRESSES,  ETC.  265 

2.  If  a  force  of  50  pounds  is  suddenly  applied  at  the  center  of 
a  2-inch  by  2  -inch  simple  timber  beam  of  6-ft.  span  what  will 
be  the  deflection  and  what  will  be  the  maximum  stress  developed  ? 

bd3  _  2  X8      4 

L    —  -    —  ——  —   =   —  j  C   =  I. 

12  12  3 

The  deflection  and  stress  developed  are  the  same  as  those 
developed  by  twice  the  static  load,  or  100  pounds. 

3  ••*>"*• 


f       Me       100  X  72  X  i  X  3 
/i.-  —  -   -          '  ^  =  1350  lb.  per  sq.  in. 

i  4X4 

3.  If  the  weight  in  Example  No.  2  falls  i  inch  before  impinging 
on  the  beam  what  stress  will  be  developed  and  what  will  be  the 
maximum  deflection? 


Ai  =  .194  +  A2  X.IQ4  X  i  +.04  =  .194  +  .65  =  .844  inch. 
The  stress  developed  is  proportional  to  the  deflection  and  is 

844 
/i  =  1350  X  —^  =  2920  pounds  per  square  inch. 

Or  the  stress  is  the  same  as  that  developed  by  a  static  load  of 


194 


218  X  72  X  3 

--  -  -  -  =  2920  pounds  per  square  inch. 
4X4 


PROBLEMS 

i.  What  is  the  resilience  stored  in  a  cubic  inch  of  the  follow- 
ing materials  when  the  stress  is  at  the  elastic  limit  (modulus  of 
resilience)?  (a)  Wrought  iron,  (b)  Structural  steel. 

Ans.    12. $  in.-lb.;  20.4  in.-lb. 


266  REPEATED   STRESSES,   ETC.         [CHAP.  XVI 

2.  What  horse  power  is  required  to  stress  a  structural  steel 
rod  2  inches  in  diameter  and  6  feet  long  from  zero  to  the  elastic 
limit  120  times  per  minute? 

3.  Solve  Problem  No.  2  if  the  stress  is  carried  from  one-half 
the  elastic  limit  to  the  elastic  limit  each  time. 

4.  If  a  load  of  2000  pounds  is  suddenly  applied  to  the  end  of  a 
steel  rod  3  feet  long  and  1.5  inches  in  diameter,  what  will  be  the 
deformation  and  the  unit-stress  developed? 

5.  If  the  load  in  Problem  No.  4  is  moving  horizontally  with 
a  velocity  of  5  feet  per  second  at  the  instant  of  impinging  on  the 
rod,  what  deformation  and  unit-stress  will  be  developed  ? 

6.  If  the  load  in  Problem  No.  4  is  falling  with  a  velocity  of 
5  feet  per  second  at  the  instant  of  impinging  on  the  rod,  what  will 
be  the  deformation  and  the  unit-stress  developed  ? 

7.  What  is  the  work  required  to  deflect  a  2 -inch  by  4-inch  timber 
beam  of  8-ft.  span  by  a  central  load  that  will  produce  a  maximum 
stress  equal  to  1 200  pounds  per  square  inch  ?     Solve  this  problem 
for  both  cases,  when  the  beam  is  on  the  edge  and  when  it  is 
lying  flat. 

8.  If  a  load  of  2  tons  falls  through  a  distance  of  ^  foot,  and 
strikes  at  the  center  of  a  io-in.,  25-lb.  I-beam  of  i6-ft.  span,  what 
deflection  and  stress  will  be  developed  ? 

9.  A  structural  steel  rod  is  required  to  support  a  suddenly 
applied  load  of  10,000  pounds.    What  is  the  minimum  diameter 
of  the  rod  if  permanent  set  is  avoided? 


APPENDIX   A 

CENTROIDS  AND  MOMENTS  OF  INERTIA  OF  AREAS 

AI.  Such  expressions  as  S  ay  and  S  ay2  will  occur  in 
finding  the  stresses  developed  in  beams  under  load, 
where  a  is  an  element  of  area  and  y  is  the  distance  of 
that  element  from  a  reference  line  or  axis.  It  is  neces- 
sary to  be  able  to  evaluate  these  expressions  for  the 
various  shapes  of  cross  sections  found  in  beams. 

A2.  CENTROIDS  OF  AREAS.  The  centroid  of  an  area 
is  the  point  at  which  a  very  thin  homogeneous  plate 


FIG. 


of  the  shape  of  the  area  would  balance:  it  is  the  point 
at  which,  if  the  area  were  concentrated,  its  moment 
about  any  axis  would  be  equal  to  the  moment  of  the 
area  as  originally  distributed.  Calling  y  the  distance 
from  the  ^T-axis  to  the  centroid  of  the  area  A,  a  an  element 

267 


268  APPENDIX  [App.  A 

of  the  area,  and  y  the  distance  of  that  element  from 
the  X-axis : 

Ay  =  2  ay, 


And  calling  x  the  distance  of  the  centroid  of  the  area 
from  the  F-axis,  and  x  the  distance  of  an  element  of  the 
area  from  that  axis, 

~      S  ax 

Ji     . 

A 

The  axes  may  be  chosen  arbitrarily  (Fig.  AI).  For 
solids  the  term  "centroid"  is  synonymous  with  "center 
of  gravity,"  and  the  latter  term  is  also  frequently  used 
with  areas. 

A3.  AXIS  OF  SYMMETRY.  If  a  straight  line  can  be 
drawn  through  an  area  dividing  it  into  two  exactly 
similar  halves,  that  line  is  called  an  axis  of  symmetry, 
and  an  area  that  can  be  divided  in  this  manner  is  called 
a  symmetrical  area.  The  areas  shown  in  Fig.  A2  are 


FIG.  A2. 

symmetrical  areas  and  the  axes  shown  are  axes  of  sym- 
metry. If  there  is  an  axis  of  symmetry  in  an  area,  the 
centroid  is  located  on  that  axis.  This  fact  simplifies  the 
solution  for  locating  the  centroids  of  a  large  number  of 
areas. 

A4.   CENTROID  OF  A  TRIANGLE.     Imagine  the  triangle 
to  be  made  up  of  a  large  number  of  strips  of  very  small 


ART.  AS]  CENTROID   OF  A  SECTOR  269 

width  parallel  to  the  base.  Each  strip  may  be  considered 
an  element  of  the  area.  The  centroid  of  any  strip  CD, 
Fig.  AS,  is  at  the  middle  point  of  its  length.  The  cen- 
troids  of  all  the  other  strips  parallel  to  the  base  come  at 
the  middle  of  their  lengths.  The  line  joining  the  centers 
of  all  these  strips  is  a  straight  line  and  is  called  a  median. 


FIG.  A3. 

The  centroid  of  the  entire  triangle  falls  on  the  median 
AB,  since  the  centroids  of  all  its  elements  fall  on  that 
line.  If  the  triangle  is  considered  as  being  made  up  of 
strips  parallel  to  another  side  it  is  shown  by  the  same 
reasoning  that  the  centroid  of  the  triangle  lies  on  another 
median.  Therefore,  the  centroid  of  a  triangle  is  at  the 
intersection  of  the  medians,  which  is  at  a  distance  of  one- 
third  the  altitude  from  the  base. 

A5.  CENTROID  OF  A  SECTOR  OF  A  CIRCULAR  AREA. 
The  centroid  of  a  circular  sector  may  be  located  in  the 
following  manner:  Let  the  angle  at  the  center  subtended 
by  the  radii  be  2  a,  and  r  be  the  radius  (Fig.  A4).  Take 
the  Jf-axis  as  the  axis  of  symmetry.  Then  ~y  =  o.  Con- 
sider the  sector  as  being  made  up  of  a  great  number  of 
triangular  elements,  as  OAB.  The  distance  of  the  cen- 
troid of  the  triangle  from  0  is  f  r,  and  the  distance  from 
the  F-axis  or  x  is  f  r  cos  6.  Draw  the  arc  CED  with 
radius  equal  to  f  r.  The  centroids  of  all  elements  of  the 
sector  fall  on  this  arc.  The  total  area  of  the  sector  may 


270 


APPENDIX 


[App.  A 


be  represented  by  the  length  of  the  arc  CED,  and  the 
area  OAB  may  be  represented  in  the  same  way  by  the 

arc  GF.  The  centroid 
of  the  sector  would  be 
the  same  as  if  the  en- 
tire area  were  concen- 
trated on  the  arc  CED. 
Draw  GH  perpendicu- 
lar to  the  X-axis  and 
HF  parallel  to  that 
axis.  Since  FG  is  very 
small  it  may  be  taken 
as  a  straight  line. 
The  angle  FGH  equals 
6.  The  -moment  of  the 
area  of  the  triangle 
OAB  about  the  F- 
axis  is  x  (GF)  = 
f  rcos0(G70,andthe 
moment  of  the  total 
sector  equals  the  sum- 
mation of  all  such  expressions,  and  is  2  §  r  cos  6  (GF) . 
This  sum  divided  by  the  total  area  gives  x.  The  area 
of  the  sector  equals  CED,  which  equals  f  r  2  a  =  f  ra, 
-  =  2%rcos6(GF)  =  2  (GF)  cos  0 _  g_(gg)  m 

%ra  2  a  2  a 

(GH) ,  and  the  summation  of  all  such  lengths 


FIG.  A4. 


(GF)  cos  6 
equals 

(CD)  =  2-f  rsina 


r  sn  a, 


_  _  2  (GF)  cos  0  _  |  sin  a  _  2     sin  a 


=  -  r 
2  a  2a          3         a 

The  angle  a  must  be  expressed  in  radians  in  applying  this 
formula. 

7T  *  4.  7" 

When  a  =  -  the  sector  is  a  semicircle  and  x  —  —  • 
2  3^ 


ART.  A6]       CENTROIDS  OF  COMPOSITE  AREAS 


271 


•  A6.  CENTROIDS  OF  COMPOSITE  AREAS.    For  such 

sections  as  often  occur  in  practice,  where  they  are  built  up 
of  several  different  parts,  or  when  the  area  may  be  divided 
into  simpler  areas,  the  centroid  of  the  area  can  be  ob- 
tained by  applying  the  fundamental  formula  y  =  — -p- 

to  the  section.  The  method  is  most  readily  understood 
from  an  example.  Let  it  be  required  to  locate  the  cen- 
troid of  the  channel  section  shown  in  Fig.  A5.  Divide 


A" 


tfL. 


FIG.  A6. 

the  section   into   three   rectangles  A,  B,  and   C.     The 
following  tabulated  values  are  then  found: 


Part. 

Area  =  a. 

y 

ay 

A 

4X-4=  1-6 

2.O 

3.200 

B 

4X.4  =  1-6 

2.O 

3.200 

C 

5.2  X-4  =  2.08 

0.2 

0.416 

A  =  5.28  square  inches, 
6.816 


2ay=  6.816, 


y  = 


5-28 


=  1.29  inch. 


Since  the  F-axis  is  an  axis  of  symmetry  x  =  o. 

Another  method  easily  applied  for  certain  sections 
results  from  subtracting  moments.  The  solution  of  the 
above  example  by  this  method  is  to  consider  the  whole 


272 


APPENDIX 


[App.  A 


rectangle  4  inches  by  6  inches  with  the  rectangle  3.6 
inches  by  5.2  inches  cut  away  from  the  top  as  indicated 


FIG.  A6. 
in  Fig.  A6.     The  following  table  is  then  obtained : 


Part. 

Area  =  a. 

y 

ay 

A 

6X4       =24.00 

2 

48.000 

B 

5.2  X  3-6  =  18.72 

2  .  2 

41  .  184 

I 

A  =  5.28,   S  ay  =  6.816,   y  =  -^-—r  =  1.29  inches. 

A7.  MOMENT  OF  INERTIA.    The  moment  of  inertia  of 
an  area  is  the  summation  of  the  products  obtained  by 


FIG.  A7. 

multiplying  each  elementary  part  of  the  area  by  the 
square  of  its  distance  from  an  axis.     The  axis  taken  is 


ART.  A9]  POLAR  MOMENT  OF  INERTIA  273 

called  the  inertia  axis.     Thus  the  moment  of  inertia  of 
an  area  shown  in  Fig.  A?  with  respect  to  the  X-axis  is: 

Ix  =  2  ay2, 
and  the  moment  of  inertia  with  respect  to  the  F-axis  is 

Iy  =  2  ax2. 

These  expressions  are  for  the  moment  of  inertia  of  the 
area  about  axes  in  the  plane  of  the  area. 

Since  the  moment  of  inertia  is  the  product  of  an  area 
and  a  length  squared,  the  units  in  which  it  is  expressed 
are  L2  X  L2  =  L4,  a  length  to  the  fourth  power. 

A8.  THE  RADIUS  OF  GYRATION.  The  radius  of  gyra- 
tion with  respect  to  an  axis  is  denned  as  the  square  root 
of  the  quotient  obtained  by  dividing  the  moment  of 
inertia  of  the  area  with  respect  to  the  same  axis  by  the 
area.  Thus,  if  /  is  the  moment  of  inertia  and  A  is  the 
area,  the  radius  of  gyration  is 


v/i 


It  is  seen  that  the  radius  of  gyration  gives  the  position 
for  which  a  concentration  of  the  area  would  give  the 
same  moment  of  inertia  as  is  found  for  the  distributed 
area.  The  value  of  r  should  not  be  confused  with  the 
distance  to  the  centroid  of  the  area. 

A9.  POLAR  MOMENT  OF  INERTIA.    THE  RELATION 

BETWEEN  THE  POLAR  MOMENT  OF  INERTIA  AND  Ix 
AND  Iy.  The  moment  of  inertia  of  an  area  about  an  axis 
perpendicular  to  the  plane  of  the  area  is  the  polar  mo- 
ment of  inertia  and  is  obtained  by  taking  the  sum  of  the 
products  formed  by  multiplying  each  element  of  the  area 
by  the  square  of  its  perpendicular  distance  to  the  axis. 
If  the  axis  is  perpendicular  to  the  plane  at  0  in  Fig.  AS, 
the  distance  to  an  element  is 

p  =  Vy2  +  x2. 


274 


APPENDIX 


[App.  A 


The  polar  moment  of  inertia  of  the  area  equals  2  ap2.    If  / 
is  the  polar  moment  of  inertia  of  the  area  about  that  axis, 

j  =  2  ap2  =  2  a  (y2  +  x2)  =  S  ay2  +  S  ax*, 


since  S  a/  =  7*  and  2  ax2  =  Iv. 


A"  A'" 


=  r  ,= 


2  +  r2. 


FIG.  A8. 


The  polar  moment  of  inertia 
of  a  plane  area  about  an  axis 
perpendicular  to  the  plane  of 
the  area  equals  the  sum  of 
the  moments  of  inertia  of  the 
area  about  two  rectangular 
axes  in  the  plane  of  the  area 
intersecting  the  given  axis. 
This  is  the  relation  between 
the  moments  of  inertia  about 
three  mutually  perpendicular 
axes,  two  of  which  lie  in  the 
plane  of  the  area. 


Aw.  RELATION   BETWEEN   MOMENTS   OF  INERTIA 
ABOUT  PARALLEL  AXES  IN  THE  PLANE  OF  THE  AREA.  In 

Fig.  A9  let  0  be  the  centroid  of  the  area,  /  the  moment 
of  inertia  of  the  area  about  the  J^f-axis,  and  I'  the  moment 
of  inertia  of  the  area  about  the  Jf'-axis  at  a  distance  d 
from  the  centroidal  axis.  Then 

/'  =  S  ay2  +  2  ad2  +  2  d  S  ay, 
2  ay2  =  I,    S  aJ2  =  yl^2,    and 
2  d  2  03;  =  2  <14y  =  o,    since   y  =  o, 

ft  T  /I   j2 

.-./'  =  7+  A'/.    '    -1 


ART.  An] 


I  OF  A  PARALLELOGRAM 


275 


The  moment  of  inertia  of  an  area  about  an  axis  parallel  to 
a  centroidal  axis  in  the  plane  of  the  area  is  equal  to  the 


FIG.  A9. 

moment  of  inertia  about  the  centroidal  axis  plus  the  area 
times  the  square  of  the  distance  between  the  two  axes. 

If  the  moment  of  inertia  of  an  area  about  any  axis  is 
given,  that  for  any  other  parallel  axis  can  be  obtained. 
First,  the  moment  of  inertia  about  the  centroidal  axis 
must  be  obtained  by  the  formula  7  =  /'  —  Ad  .  Second, 
the  moment  of  inertia  about  the  parallel  axis  can  be 
obtained  by  the  use  of  the  formula  I"  =  /+  Ad  . 


An.  THE  MOMENT  OF  INERTIA  OF  A  PARALLELO- 
GRAM ABOUT  A  CENTROIDAL  Axis  IN  THE  PLANE  OF 
THE  AREA.  The  inertia  axis  is  taken  parallel  to  opposite 
sides,  b  is  the  breadth  of  the  parallelogram  and  d  is  the 
depth  perpendicular  to  the  chosen  axis,  Fig.  AID.  Let  the 
area  be  divided  into  a  large  number  n,  of  equal  strips  par- 
allel to  the  axis,  each  strip  being  taken  so  small  in  width 
that  it  is  an  element  of  the  given  area.  The  width  of 

each  strip  is  -  ,  and  the  area  of  each  strip  is 

d, 

a  =  -b. 
n 

Let  the  strip  shown  be  the  pth  one  from  the  axis  in  which 
p  is  any  number  up  to  -  ,  then  the  distance  from  the  axis 


276  APPENDIX 

to  the  element  of  the  area  shown  is 


[App.  A 


The  moment  of  inertia  of  the  parallelogram  about  the 
centroidal  axis  is 


FIG.  Aw. 

To  obtain  the  moment  of  inertia  for  the  area  of  the  paral- 
lelogram above  the  axis,  p  must  represent  all  numbers  up 

to  -  .     The  same  is  true  for  the  area  below  the  axis, 
therefore, 


ART.  An]  I  OF  A  PARALLELOGRAM,  277 

From  algebra, 


_  n3  +  3  n2  -f  2  n 
~^~ 

+  3  n* 


_ 
=  12  > 


The  greater  w  is  made  the  more  nearly  is  the  true  value 
for  /  obtained,  and  when  n  becomes  infinitely  great  the 
exact  value  of  the  moment  of  inertia  is  obtained.  For 

^  2 

this  condition  -  and  —  2  become  zero, 

-  _  bd* 

"    12* 

The  radius  of  gyration  with  respect  to  the  centroidal 

axis  is 


II         /bd*      ,.         /d2 

=  \/-r  =  I/ '-  bd  =  I/—  = 

V  A       V   12  V  12 


2 

The  moment  of  inertia  of  a  parallelogram  about  one 
of  its  sides  is  often  needed,  and  by  the  application  of  the 
formula 

Ir=I  +  Ad\ 


The  corresponding  radius  of  gyration  is 

d 

r  =  ^' 

The  rectangle  is  the  usual  form  of  parallelogram  for 
which  the  moment  of  inertia  is  needed. 


278 


APPENDIX 


[App.  A 


FIG.  An. 


x' 


FIG,  A12, 


ART.  Ai3]  I  OF  A  CIRCULAR  AREA  279 

Ai2.  THE  MOMENT  OF  INERTIA  OF  A  TRIANGLE 
ABOUT  ITS  CENTROIDAL  AXIS.     The  moment  of  inertia 

bd? 
of  the  parallelogram  about  the  J^'-axis,  Fig.  An,  is  —  • 

The  rectangle  may  be  considered  as  being  made  up  of 
two  triangles  ABC  and  ABD,  both  equal  and  similar. 
Consequently  the  moments  of  inertia  of  the  two  tri- 
angles about  the  ^'-axis  are  equal.  Therefore,  the 
moment  of  inertia  of  one  of  the  triangles  about  that 

axisis  r,      U*  U* 

Ix'  =  --   -T-  2  =  —  • 
12  24 

This  axis  is  at  the  distance  -  from  the  centroidal  axis 

o 

of  the  triangle  (Fig.  Ai2), 

/.    Jx  =  Ixr  -  Ad\ 

j   _  bd3      bd       d*  _  bd3 
J.  x  —  X     ~^  —      ~^  * 

24          2          36         36 

The  corresponding  value  of  the  radius  of  gyration  is 

d 

'  =  3V2; 

b  is  the  base  of  the  triangle  and  d  is  the  altitude. 
Ais.  THE  MOMENT  OF  INERTIA  OF  A  CIRCULAR  AREA. 

Let  d  be  the  diameter  of  the  circle.  Let  the  area  be 
divided  into  a  great  number,  n,  of  elementary  annular 
strips  concentric  with  the  entire  area,  Fig.  Ai3.  The 

width  of  each  strip  will  be  ---     Let  the  strip  shown  in 

2  n 

Fig.  A  13  be  the  pih  strip  from  the  center,  then  the  radius 
of  this  strip  is 

2  =  M. 

2  n 
The  area  of  the  element  is 

*d       d       -K& 


n        2n 


280 


APPENDIX 


[APP.  A 


The  polar  moment  of  inertia  of  the  entire  area  about  the 
axis  perpendicular  to  the  area  at  O  is 


in  which  p  represents  all  numbers  up  to  n. 

=  S  (I3  +  23  +  -  -  '  +  p*  +  •  • 


FIG.  AM. 
From  algebra, 

2  (l3  +  23  +  •    •    '   +  P*  +  •    •    •   +  W2) 

n2(n  +  i)2      w4  +  2w3  + w2 

=  —  -  = —  > 

4  4 

...     /  =  ^4/rc4  +  2rc3  +  rc2\  =  7rJ4/I    [      T      [      r 


Since  w  should  be  made  infinitely  great  to  obtain  the  true 

-  and 
2  n          4  n 


moment  of  inertia,  -  -  and  —  r  reduce  to  zero,  and 

2 


32 


ART. 


I   OF   COMPOSITE  AREAS 


28l 


From  Art.  A9  J  =  Ix  -(-  Iy  and  from  the  symmetry  of 
the  figure  Ix  =  Iy, 

-       -       J      *# 


rx 


TV   =  - 


Ix  is  the  moment  of  inertia  of  the  circle  about  a  diameter. 


9.4" 


Ai4.  MOMENT  OF  INERTIA  OF  COMPOSITE  AREAS. 

In  order  to  obtain  the  moment  of  inertia  of  a  built-up 

section  (composite  area)  for  a  given  axis  the  area  should 

be  divided  into  its  simpler  parts,  and  the  moment  of 

inertia  of  each  part  with  respect  to  the  given  axis  ob- 

tained.    The  moment  of  inertia  of 

the  entire  area  with  respect  to  the 

axis  equals  the  sum  of  the  moments 

of  inertia  of  its  component  parts 

with  respect  to  the  axis.     The  ap- 

plication can  be  understood  by  an 

example. 

Let  it  be  required  to  determine 
the  moment  of  inertia  and  the 
radius  of  gyration  of  a  T-section 
8  inches  by  9.4  inches  by  0.4  inch 
with  respect  to  a  centroidal  axis  parallel  to  the  flange  of 
the  T,  Fig.  A14. 

By  taking  moments  about  the  X-axis, 

_       8  X  .4  X  9.2  +  9  X  .4  X  4.5 

y  '-  8  X  4  +  9X4  =  6'71  mches- 

For  the  part  A  , 
bd*  : 


—  X 


FIG.  A] 


12 


8  X  .064 

12 


X  2.492 


.04  +  19.84  =  19.88  inches. 


282 

For  the  part  B, 


APPENDIX  [App.  A 


+  3-6  X  2.2I2  =  24.3  +  17.58 

=  41.88  inches4. 

Therefore,  the  moment  of  inertia  of  the  section  about  the 
centroidal  axis  is 

7  =  19.88  -f  41.88  =  61.76  inches4. 


3.01  inches. 


6.8 


EXAMPLE 

i.   Determine  the  moment  of  inertia  and  the  radius  of  gyration 
with  respect  to  the  axis  through  the  base  and  the  centroidal  axis, 
of  a  channel  section,  4  inches  by  6 
inches  by  0.4  inch,  Fig.  Ai6. 

The  moment  of  inertia  can  be  ob- 
tained for  either  axis  and  then  trans- 
ferred to  the  other,  or  it  can  be 
obtained  for  each  one  independently. 
That  for  the  axis  through  the  base 
will  be  obtained,  then  the  transfer  to 
the  centroidal  axis  made. 


0.4" 

1.29" 

A 

B 

± 

c 

4- 

6-  

-> 

FIG.  Ai6. 


Part. 

Area. 

J 

d 

A$ 

bd* 

12 

I 

A 

B 
C 

1.6 
1.6 
2.08 

2.O 
2.O 
O.  2 

2.O 
2.O 
0.  2 

6.4 
6.4 
.083 

2-133 
2-133 
.027 

8-533 
8-533 
.  II 

5-28 

12.883 

4-2Q3 

17.176 

Ix  =  17.176  inches4,  say  17.18,  and  r*  =  \          n  =  1.8  inches. 

>    5.28 

d  =  y  =  1.29  inches.     (Ex.  p.  271.) 

1  =  1-  Ad*  =  17.18  -  5.28  X  i.2Q2  =  17.18  -  8.77 

=  8.41  inches4  and  r  =  \/  —  ^  =  1.26  inches. 
V  5>2° 


PROB.] 


APPENDIX 


283 


PROBLEMS 

1.  Find  the  distance  of  the  centroid  of  a  trapezoid  with  one 
base  a,  the  other  b,  and  the  altitude  h,  from  the  base  whose 
length  is  a. 

An,.  y=^\h. 
3<*  +  3& 

2.  Determine  the  moment  of  inertia  and  radius  of  gyration 
with  respect  to  the  Jf-axis  arid  with  respect  to  the  F-axis  passing 
through  the  centroid  of  the  area  shown  in  Fig.  A 16. 

Ans.  Ix  =  $9.3  in.4,  Iy  =  29. 8  in.4,  ry  =  1.68  in. 


:0.4' 


5"9.75*Is- 


0.4: 


JW 


LLL. 

FIG.  An. 


3.  Locate  the  centroid  and  determine  the  moment  of  inertia 
and  the  radius  of  gyration  with  respect  to  the  X  and  Y  axes 
through  the  centroid  for  the  T-section  shown  in  Fig.  An. 

Ans.   y  =  4.01  in.,/x  =  17. 4  in.4 

4.  Calculate  the  moment  of  inertia  and  radius  of  gyration  of 
a  circular  area  of  diameter  4  inches  with  respect  to  the  diameter 
and  with  respect  to  a  tangent.     Also  find  the  polar  moment  of 
inertia  with  respect  to  the  center. 

5.  A  section  is  built  up  of  two  15 -in..  33-lb.  channels  placed 
back  to  back.     What  should  be  the  distance  between  them  to 
have  the  moments  of  inertia  of  the  section  equal  with  respect 
to  the  two  rectangular  axes  passing  through  the  centroid  of  the 
section?  Ans.  9.5  in.  from  back  to  back. 


284 


APPENDIX 


[APP.  A 


6.  Find  the  moment  of  inertia  and  radius  of  gyration  with 
respect  to  the  centroidal  X  and  Y  axes  of  the  I-section  shown 
in  Fig.  Ai8. 

7.  A  girder  is  built  up  of  four  6-inch  by  6-inch  by  i-inch 


FIG.  A  is. 


FIG.  A 19. 


FIG.  A20. 


angles  and  a  30-inch  by  i-inch  plate;  determine  the  moment  of 
inertia  of  the  section  with  respect  to  the  centroidal  X  and  Y 
axes.  See  Fig.  A19. 

8.  Locate  the  centroid  of  the  shaded  area  shown  in  Fig.  A20, 
and  find  the  moment  of  inertia  with  respect  to  the  axis  parallel 
to  a  side  and  passing  through  the  centroid. 

Ans.  x=y  =  .776 r in., IX=IV  =  .1368 r* in.4, Ix=Jy  =  .0075 r4 in.4 

Prove  that  the  moment  of  inertia  of  each  of  the  following  areas 
about  the  centroidal  axis  and  the  corresponding  radius  of  gyration  are 
as  given : 


9. 


/*  = 


bd3  -  Mi3 


12 


bd3  -  Mi3 
12  (bd  -MO 


PROB.] 


APPENDIX 


Ix  = 


64 


285 


-Si :: 


(b 


13. 


L 


_    J     bd*    -k*(b-  t) 

Tx      V 


I2[bd-h(b-£)] 


14. 


+  hf 


2  sb3  +  /i/3 


286 


APPENDIX 


[APP.  A 


TABLE   21 

PROPERTIES  OF  STANDARD  LIGHT  STEEL  I-BEAM  SECTIONS 


Axis  perpendicular  to  web. 

Axis  parallel 
to  web. 

Weight 

Section 

Depth, 

I__l_  __ 

per  foot, 

area, 

ncnes. 

Pounds. 

Sq.  in. 

I 

I/c 

r 

/ 

r 

Inches*. 

Inches3. 

Inches. 

Inches4. 

Inches. 

24 

80 

23.32 

2088.0 

174.0 

9.46 

42.86 

.36 

20 

80 

23-73 

1467.0 

146.7 

7.86 

45-81 

•39 

20 

65 

19.08 

1170.0 

117.0 

7.83 

27.86 

.21 

18 

55 

15-93 

795-6 

88.4 

7.07 

21.19 

-IS 

15 

80 

23-57 

789.1 

105.2 

5-79 

41.31 

-32 

IS 

60 

17.67 

609.0 

81.2 

5-87 

25.96 

.21 

IS 

42 

12.48 

441.8 

58.9 

5-95 

14.62 

.08 

12 

40 

11.84 

268.9 

44-8 

4-77 

13.81 

08 

12 

31* 

9.26 

215.8 

36.0 

4-83 

9-50 

.OI 

IO 

25 

7-37 

122.  1 

24.4 

4.07 

6.89 

0.97 

9 

21 

6.31 

84.9 

18.9 

3-67 

S.I6 

0.90 

8 

18 

5-33 

56-7 

14.2 

3-27 

3-78 

0.84 

7 

IS 

4-42 

36.2 

10.4 

2.86 

2.67 

0.78 

6 

I2j 

3-6i 

21.8 

7-3 

2.46 

1.85 

0.72 

5 

9f 

2-37 

12.  I 

4-8 

2.05 

1.23 

0.65 

4 

1\ 

2.21 

6.0 

3-o 

i  .64 

0.77 

0-59 

3 

sl 

I-63 

2.5 

i-7 

1.23 

0.46 

0-53 

TABLE   22 

PROPERTIES  OP  STANDARD  LIGHT  STEEL  CHANNEL  SECTIONS 


Axis  perpendicular  to  web. 

Axis  parallel 
to  web. 

Depth, 
Inches. 

Weight 
per  foot, 
Pounds. 

Section 
area, 
Sq.  in. 

7 

r 

I 

r 

y 

Inches4. 

Inches. 

Inches4. 

Inches. 

Inches. 

IS 

33 

9.90 

312.6 

5-62 

8.23 

0.91 

0.79 

12 

20* 

6.03 

128.1 

4.61 

3-91 

0.81 

0.70 

IO 

IS 

4-46 

66.9 

3-87 

2.30 

0.72 

0.64 

9 

i3i 

3-89 

47-3 

3-49 

1.77 

0.67 

0.61 

8 

"i 

3-35 

32.3 

3.10 

1-33 

0.63 

0.58 

7 

9i 

2-85 

21  .1 

2.72 

0.98 

o-59 

o-SS 

6 

8 

2.38 

13.0 

2-34 

0.70 

0-54 

0.52 

5 

6| 

i-95 

7-4 

i-95 

0.48 

0.50 

0.49 

4 

si 

i-SS 

3-6 

i-S6 

0.32 

o-4S 

0.46 

3 

4 

1.19 

1.6 

1.17 

O.2O 

0.41 

0.44 

TABLES 


287 


TABLE  i 

WEIGHTS  OF  VARIOUS  MATERIALS  USED  IN  CONSTRUCTION 


Material. 

Weight, 
lb.  per  cu.  ft. 

Material. 

Weight, 
lb.  per  cu.  ft. 

Timber  
Cast  iron  
Wrought  iron  . 
Steel  

25  to  45 

450 
480 

400 

Sandstone  
Granite  
Marble  
Slate  

ISO 

170 
170 

I7C 

Brass  
Copper,  Bronze 
Aluminum  .... 

515 
550 
1  60 

Terra  cotta,  facing.  .  .  . 
Terra  cotta,  fireproof- 
ing  .  . 

no 
5° 

Brick  

100  to  150 

Book  tile  

60 

Limestone.  .  .  . 

165 

Concrete  

150 

TABLE   2 

ULTIMATE  TENSILE  STRENGTH  AND   ULTIMATE  ELONGATION 
OF  MATERIALS 


Material. 

Ultimate  tensile  strength, 
lb.  per  sq.  in. 

Ultimate  elongation, 
per  cent. 

Timber  

6,000  to  10,000 

I    ^ 

Cast  iron 

2O  OOO 

Wrought  iron  

5O,OOO 

•2Q     O 

Structural  steel 

60  ooo 

2C    n  tO  ?O    O 

Steel  wire  

60  ooo  to  250  ooo 

10  o  to  25  o 

288 


STRENGTH   OF  MATERIALS 


TABLE  3 
ULTIMATE  COMPRESSIVE  STRENGTH  OF  MATERIALS 


Material. 

Ultimate  compressive 
strength,  Ib.  per  sq.  in. 

Timber 

7  OOO 

Cast  iron 

OO  OOO 

Brick  .                  

6,OOO 

Brick  masonry  
Rich  concrete  
Stone  

1,500 
2,500 
IO,OOO 

TABLE  4 
ULTIMATE  SHEARING  STRENGTH  OF  MATERIALS 


Material. 

Ultimate  shearing 
strength,  Ib.  per  sq.  in. 

Timber: 
Along  grain 

4.OO 

Across  grain 

3.OOO 

Cast  iron 

2O,OOO 

Wrought  iron  
Structural  steel  
Rivet  steel  

40,000 
50,000 
45,000 

TABLE  5 
ELASTIC  LIMIT  OF  WROUGHT  IRON  AND  STEEL 


Material. 

Elastic  limit, 
Ib.  per  sq.  in. 

Wrought  iron  
Structural  steel  
Hard  steel 

25,000 
35.000 

*\O  OOO 

TABLES 


289 


TABLE  6 
MODULUS  OF  ELASTICITY 


Material. 

Modulus  of  elasticity, 
Ib.  per  sq.  in. 

Timber         

I,5OO,OOO 

Cast  iron  

15,000,000 

Wrought  iron  

25,000,000 

Steel 

30,000,000 

TABLE  7 
SHEARING  MODULUS  OF  ELASTICITY 


Material. 

Shearing  modulus  of 
elasticity,  Ib.  per  sq.  in. 

Timber,  across  grain.  .  .  . 
Cast  iron  

4OO,OOO 
6,OOO,OOO 

Wrought  iron 

IO,OOO,OOO 

Steel 

I2,OOO,OOO 

2  go 


STRENGTH  OF  MATERIALS 


TABLE  8 

SAFE  WORKING  STRESSES  IN  POUNDS  PER   SQUARE  INCH  FOR 
STEADY  LOADS 


Material. 

Tension. 

Shear. 

Compression. 

Bending 
(fiber). 

Perpendic- 
ular to 
grain. 

Parallel 
to  grain. 

Timber: 
Cedar,  white 

800 
600 
I,OOO 
I,2OO 
I,OOO 
800 
800 
I,OOO 

800 
900 

1,000 

1,000 

IOO 
IOO 

240 

IOO 
200 

80 

120 
200 

160 

•    1  60 

200 
I2S 
IOO 
IOO 

IOO 
IOO 

2,500 

9.500 

10,000 

(    8,000 

(  10,000 

1  80 
1  80 
300 
300 
340 
1  80 
240 
800 
500 
500 
600 
240 
20O 
20O 

200 

200 

1,100 
I,  IOO 
I,2OO 
1,  6oO 
1,300 
I,OOO 
I,2OO 
1,  800 
I,40O 
I,20O 
1,750 

,400 

,000 
,200 

,200 
,200 

I,OOO 

1,000 

I,2OO 
I,2OO 
I,  IOO 
800 
1,300 
1,  800 
I,2OO 
1,200 
I,4OO 
I,2OO 
I,OOO 
I,  IOO 

I,OOO 
I,OOO 

6,000 

I  2  ,OOO 

16,000 

Cypress 

Elm 

Fir,  Washington  .... 
Gum        .    .         ... 

Hemlock  

Larch           .        ... 

Maple,  sugar  (hard). 
Maple  (average)  
Oak,  red  
white  

Pine,  longleaf  
loblolly  
shortleaf  

yellow,    (Ark., 
etc.)  
Spruce      

800 
800 

3,000 
12,000 
15,000 

Cast  iron       

12,000 
I  2  ,000 
12,000 

18,000  (Bearing) 
no 

250 
350 

Wrought  iron          .... 

Steel,  structural  
rivet  

Brickwork  (in  lime)  .  .  . 
Brickwork  (in  Portland 
cement)  

Concrete     (Portland 
cement) 

TABLES 


291 


TABLE  9 
COEFFICIENTS  OF  EXPANSION  PER  DEGREE  FAHR. 


Material. 

Coefficient  of  expansion. 

Masonry  

.  0000050 

Cast  iron  

.0000062 

Wrought  iron  
Steel 

.0000067 
.0000065 

TABLE  10 
VALUES  OF  K,  K,'  AND   C  FOR  PIPES  UNDER  EXTERIOR  PRESSURE 

(For  use  in  the  Carman  and  Carr  Formula.) 


Material. 

K. 

K'. 

C. 

Cold-drawn  seamless  steel  
Brass  
Lap-  welded  steel 

50,200,000 
25,150,000 

95.520 

93.365 
83,270 

2,090 
2,474 

I,O2< 

TABLE   ii 
EFFICIENCY  OF  JOINTS 


Kind  of  joint. 

Efficiency, 
per  cent. 

Single-riveted  lap  joint  

^0-65 

Double-riveted  lap  joint  
Single-riveted  butt  joint  

65-75 
6^-75 

Double-riveted  butt  joint  

70-80 

2Q2 


STRENGTH  OF  MATERIALS 


TABLE  12 
RELATIVE  STRENGTHS  IN  SHEAR  AND  BENDING 


Kind  of  beam. 

Maximum 
vertical 
shear. 

Maximum 
bending 
moment 

Relative 
strength  in 
shear. 

Relative 

strength  in 
bending. 

I                          I 

TT77 

;                   I                 >P 

W 

I 

I 

it 

Wl 

j                 w/(*/in.                | 

W 

2 

I 

2- 

rfcf 

W 

Wl 

r                f 

2 

4 

4 

w/mn.       | 

W 

Wl 

2 

8 

f                           f 

TABLE  13 
MODULUS  OF  RUPTURE 


Material. 

Modulus  of  rupture 
Ib.  per  sq.  in. 

Timber 

7000  to  oooo 

Cast  iron.  . 

3<\  OOO 

TABLES 


293 


TABLE  14 

MAXIMUM  MOMENTS  AND  MAXIMUM  DEFLECTIONS 


Kind  of  beam. 

Maximum 
moment 

Maximum 
deflection 
A. 

i 

—  PFJ 

IF/3 

^ 

Y' 

3  El 

W/^/in. 

fy 

Wl 

Wl3 

1                              IW 

t  ^=3 

// 

2 

BEI 

r 

' 

Wl 

Wl3 

4 

48  El 

Y//  1  (/in. 

Wl 

SWP 

r 

1 

8 
.JW 

384  £/ 
Tf/3 

i 

±8 

192^7 

W/i  $/in 

W7/3 

/    / 

12     ^    2A 

78/1  F7 

304  /ii 

294 


STRENGTH  OF  MATERIALS 


TABLE  15 

LOAD  TO  CAUSE  A  GIVEN  MAXIMUM  STRESS  AND  A  GIVEN 
MAXIMUM  DEFLECTION 


Kind  of  beam. 

Load  tF  to  cause 
stress  /. 

Load  W  to  cause 
deflection  A 

w 

^ 

f                              i 

8 

,  e. 

3     -7A 

/ 

P 

d 

3      /3  ^ 

V!/l  y  in. 

^ 

,  A 

8     £/A 

J 

k—  </*-4w 

y 
// 

<:/ 

/«  A 

f 

^ 

48fA 

vy/^in. 

8    £ 

76*    f  A 

'/. 
g 

r 

| 

c/ 
,    // 

/3    " 

JBf. 

•/, 

\ 

1 

8   ,7 

rr 

192      ^3  A 

r. 

W/jyin. 

I 

12   ^ 

384    f  A 

'/ 

General  type. 

J 

^ 

TABLES  295 


TABLE   16 

COEFFICIENTS  OF   wl  FOR  THE  VERTICAL  SHEAR  AT  THE 
SUPPORTS  OF  CONTINUOUS  BEAMS 

No. 
—  4"&  Spans 

QI  o g  *^ 


-H. — £ — i — 4 

l£ jollo"  iollo" To" 

000  O  — 


1JL  -HIM  13'13  _15|J 

2&        28|28        28;28        28 U 


t< — i 

1 


4ii_5^    53_!J3_    _JI|Ji.    __55.!_CL      _  JL 

104' i  104 104;  104 104!  H)4 104;  104 101, 

° » ° 2 2  O 


TABLE   17 

COEFFICIENTS  OF  -w/2  FOR  THE  BENDING  MOMENT  AT 
THE  SUPPORTS  OF  CONTINUOUS  BEAMS 

No. 
Spans 


0                          /8 

0 

f  7  ?f— 

1        it 

0 

Ho 

1         ?l 

Mo 

0 

fr 

>V 

zi"  " 

0 

5/28                     2/28 

-    >)<                1 

sAs 

0 

f  —  r~ 

f  T  

—  ^  ~3 

Hs 

4/3R                       0 

i    ^ 

_        8/<04 

%4 

0 

£=¥ 


296 


STRENGTH   OF  MATERIALS 


TABLE  18 

MAXIMUM  ALLOWABLE   COMPRESSIVE   STRESS 

IN  POUNDS  PER  SQUARE  INCH,  CHICAGO 

BUILDING  ORDINANCE,   1910. 


Material. 

Compressive  stress, 
pounds  per  square  inch. 

Steel.  . 

I4,OOO 

Wrought  iron  

IO,OOO 

Cast  iron  

IO,OOO 

TABLE   19 
VALUES  OF   <£  USED  IN  RANKINE'S  FORMULA 


Conditions  of  the  ends. 

Timber. 

Cast  iron. 

Wrought 
iron. 

Steel. 

4 

4 

4 

4 

3000 
I.78 

5000 
I.78 

36,000 
I.78 

25,000 
1.78 

Tlnth  pnrlc;  fiTPrl 

3000 

I 

5000 
I 

36,000 

I 

25,000 
I 

3000 

5000 

36,000 

25,000 

TABLE  20 

TESTS  ON  WROUGHT  IRON  UNDER  REPEATED   STRESSES 
[Wohler.} 


Number  of 
applications. 

Unit-stress  producing 
rupture. 

800 
107,000 
450,000 
10,140,000 

52,800 
48,400 
39,000 
35.000 

TABLES 


297 


TABLE   21 
PROPERTIES  OF  STANDARD  LIGHT  STEEL  I-BEAM  SECTIONS 


Axis  perpendicular  to  web. 

Axis  parallel 
to  web. 

Depth, 
Inches. 

Weight 
per  foot, 
Pounds. 

Section 
area, 
Sq.  in. 

I 

I/c 

r 

7 

r 

Inches*. 

Inches3. 

Inches. 

Inches4. 

Inches. 

24 

80 

23-32 

2088.0 

174.0 

9.46 

42.86 

-36 

2O 

80 

23-73 

1467.0 

146.7 

7.86 

45-81 

•39 

2O 

65 

19.08 

1170.0 

II7.O 

7.83 

27.86 

.21 

18 

55 

15-93 

795-6 

88.4 

7.07 

21.19 

-IS 

15 

80 

23-57 

789.1 

105.2 

5-79 

41.31 

•32 

15 

60 

17.67 

609.0 

8l.2 

5-87 

25.96 

.21 

15 

42 

12.48 

441.8 

58.9 

5-95 

14.62 

.08 

12 

40 

11.84 

268.9 

44-8 

4-77 

13.81 

I    08 

12 

3*i 

9.26 

215.8 

36.0 

4-83 

9-50 

1.  01 

IO 

25 

7-37 

122.  1 

24.4 

4.07 

6.89 

0.97 

9 

21 

6.31 

84.9 

18.9 

3-67 

5-16 

0.90 

8 

18 

5-33 

56-7 

14.2 

3-27 

3-78 

0.84 

7 

15 

4.42 

36.2 

10.4 

2.86 

2.67 

0.78 

6 

12;- 

3-6i 

21.8 

7-3 

2.46 

1-85 

0.72 

5 

9i 

2-37 

12.  I 

4.8 

2.05 

1.23 

0.65 

4 

7i 

2.21 

6.0 

3-o 

1.64 

0.77 

o-59 

3 

51 

I-63 

2-5 

i-7 

1-23 

0.46 

o-53 

TABLE   22 
PROPERTIES  OF  STANDARD   LIGHT  STEEL  CHANNEL   SECTIONS 


Axis  perpendicular  to  web. 

Axis  parallel 
to  web. 

Weight 

Section 

Depth, 
Inches. 

per  foot, 
Pounds. 

area, 
Ski.  in. 

7 
Inches4. 

r 

Inches. 

7 
Inches4. 

r 

Inches. 

Inches. 

IS 

33 

9.90 

312.6 

5.62 

8.23 

0.91 

0-79 

12 

20i 

6.03 

I28.I 

4.61 

3-91 

0.81 

0.70 

10 

15 

4.46 

66.9 

3-87 

2.30 

0.72 

0.64 

9 

I3l 

3-89 

47-3 

3-49 

1.77 

0.67 

0.61 

8 

IIj 

3-35 

32.3 

3-io 

i-33 

0.63 

0.58 

7 

9f 

2-85 

21.  I 

2.72 

0.98 

°-59 

o-55 

6 

8 

2.38 

13.0 

2-34 

0.70 

0-54 

0.52 

5 

6* 

!-95 

7-4 

i-95 

0.48 

0.50 

0-49 

4 

5i 

i-55 

3-6 

1-56 

0.32 

0-45 

0.46 

3 

4 

1.19 

1.6 

1.17 

o.  20 

0.41 

0.44 

INDEX 


PAGE 

Area,  reduction  of 19 

sectional 10 

symmetrical 268 

Areas,  centroids  of 267 

centroids  of  composite 271 

moment  of  inertia  of  circular 279 

moment  of  inertia  of  composite 281 

Axes,  choice  of  coordinate 192 

Axis,  inertia 273 

neutral 87,  88 

of  symmetry 268 

Bar,  resilience  of  a,  under  direct  stress 258 

Beam 47 

cantilever 47,  65,  73,  140,  144,  147 

continuous 48,  1 78 

fixed 48 

fixed  and  supported 1 70 

fixed  at  both  ends 154,  158,  176 

forces  acting  on  a,  as  a  whole 49 

forces  acting  on  a  portion  of  a 51 

overhanging 48,  168,  170 

resilience  of  a 258 

simple 47,  65,  73,  148,  150 

Beams,  continuous 1 78 

deflection  of 135 

design  of 95 

essential  quantities  to  be  known  about 196 

hinging  points  for  continuous 185 

horizontal  shear  in 208 

investigation  of 93 

methods  of  loading 48 

of  uniform  strength 99 

radius  of  curvature  of 130 

299 


300  INDEX 

PAGE 

Beams,  relative  strength  and  stiffness  of 160 

safe  loads  for 94 

the  three  problems 93 

Bending 130 

combined  twisting  and 251 

Bending  moment 58 

diagrams 61 

diagrams  for  cantilever,  and  simple  beams 65 

relation  between  the  vertical  shear  and  the 62 

sign  and  unit  of 59 

the  maximum 64,  65,  74 

the  rate  of  change  of 63 

the  values  of 60 

values  of  the  maximum 64,  74,  164 

Brittleness 3 

Butt  joint 33,  38 

Cantilever  beam 47,  65,  73,  140,  144,  147 

Cast  iron 4 

Centroid,  of  a  triangle 268 

of  a  sector  of  a  circular  area 269 

Centroids,  of  areas 267 

of  composite  areas 271 

Chimneys 106 

Circular  area,  centroid  of  sector  of  a 269 

moment  of  inertia  of  a 279 

Closing  line 124 

Coefficients  of  expansion 27 

Columns  and  struts 221 

behavior  of,  under  load 237 

comparative  strength  of;  condition  of  the  ends 230 

eccentric  loads  on 228,  234 

long,  Euler's  formula 233 

Rankine's  formula 231 

stiffness  of 221 

the  methods  of  transmitting  loads  to 229 

the  straight-line  formula 225 

the  strength  of 222 

the  three  problems 234 

Combined  stresses 216,  251 

Comparative  strength  of  columns 230 

Composite  areas,  centroids  of 271 


INDEX  301 

PAGE 

Composite  areas,  moments  of  inertia  of 281 

Compression 12 

combined  flexure  and  tension  or 214 

Concentrated  load 48 

Concrete,  plain  and  reinforced 4 

Constant  of  integration 119,  124,  192 

determination  of 193 

Continuous  beam 48 

Contraflexure,  points  of 157 

Coordinate  axes,  choice  of 192 

Curve,  elastic 47,  130,  135 

load 55 

Cylinders,  thick,  under  interior  pressure 28 

under  exterior  pressure 29 


Dams 106 

Danger  section 64 

Dead  loads 49 

Deflection  of  beams 135 

maximum 162 

rate  of  increase  of " 137 

relation  between  the  maximum  stress  and  maximum  deflection. .  164 

Deformation 9 

Design  of  riveted  joints 39 

Detrusion 9 

Diagrams,  bending  moment 61 

free-body 1 1 

load  and  shear 54 

load,  shear,  and  moment,  for  cantilever  and  simple  beams 65 

maximum  stress 97 

relation  between  the  load  and  shear 57 

stress-deformation 15 

Direct  stresses 9 

simple  cases  of 25 

Drop  loads 264 

Ductility 3 

Eccentric  loads,  case  of,  caused  by  a  combination  of  the  weights  of 

the  material  and  lateral  pressure no 

Eccentric  loads  on  columns 228 

Eccentric  loads  on  short  prisms 107 


302  INDEX 

PAGE 
Eccentricity  of  a  load  that  will  produce  zero  stress  in  the  outside 

fiber 108 

Efficiency  of  riveted  joints 38 

Elastic  curve 47,  130,  135 

Elastic  limit 16 

Elastic  resilience 256 

Elasticity 2 

coefficient  of 17 

modulus  of 17 

Elongation 9 

ultimate 1 1 

Euler's  formula,  long  columns 233 

Expansion,  coefficients  of 27 

Factor  of  safety 21 

Fiber  stresses,  distribution  of 87 

First  integrated  curve 116 

Five  curves,  relation  between  the 137 

Flexure,  combined,  and  tension  or  compression 214 

Force 9 

Free-body  diagram , 1 1 

Funicular  polygon 121 

Girders,  plate 212 

Gyration,  radius  of 273 

Hinging  points  for  continuous  beams 185 

Hooke's  law 17 

Hoop,  stresses  in  a 26 

Horizontal  shear  in  beams 208 

Hysteresis,  mechanical 259 

Impact  loads 262 

Inertia  axis 273 

Inflection  points 15? 

Integrated  curves 116,  117 

Integration,  constant  of 119,  124,  192 

Internal  stresses 51 

Joints,  riveted 32 

boiler 32 

butt..  33.  38 


INDEX  303 

PAGE 

Joints,  compression  loads  for 38 

design  of  riveted 39 

efficiency  of  riveted 38 

kinds  of  riveted 32 

lap,  single-riveted 35 

lap,  double-riveted 36 

lap,  with  more  than  two  rows  of  rivets 37 

methods  of  failure  of  riveted 33 

pipe 32 

structural 32 

tank 32 

Kern,  the no 

effect  when  the  resultant  falls  outside  of  the in 

the  maximum  stress  when  the  line  of  action  of  the  resultant  falls 
outside  the  middle  third  for  rectangular  prisms  which  take 

no  tension 112 

Lag 260 

Lap  joint 32 

Live  loads 49 

Load,  axial 10 

case  of  eccentric,  caused  by  a  combination  of  the  weight  of  the 

material  and  lateral  pressure no 

diagrams 54 

diagrams  for  cantilever  and  simple  beams 65 

Loads,  drop 264 

eccentric,  on  columns 228 

eccentric,  on  short  prisms 107 

moving,  on  beams 74 

impact 262 

suddenly  applied 261 

Masonry 3 

brick 3 

concrete 4 

stone 3 

Mechanical  hysteresis 259 

Mechanical  properties 2 

Modulus  of  elasticity 17 

uses  of 20 

Modulus,  shearing 18 


304  INDEX 

PAGE 

Modulus  of  resilience 256 

Modulus  of  rigidity ig 

Modulus  of  rupture 97 

Modulus  of  transverse  elasticity 19 

Modulus,  section go 

Young's 17 

Moment,  assumptions  for  the  resisting 86 

bending 58 

maximum 64,  74,  162 

resisting 86 

resisting,  assumptions  for  the 86 

Moment  formula,  the . go 

Moment  of  inertia 272 

of  a  circular  area 279 

of  composite  areas 281 

of  a  parallelogram  about  a  centroidal  axis 275 

of  a  triangle  about  a  centroidal  axis 27g 

polar 273 

relation  between,  about  parallel  axes 274 

Moments,  the  theorem  of  three 181,  205 

Moving  concentrated  loads  on  a  beam 74 

Neutral  axis 87 

position  of  the 88 

Neutral  surface 87,  132 

position  of  the 88 

slope  of  the 132 

Non-uniform  loads 4g 

Overhanging  beam 48,  168,  1 70 

Parallelogram,  moment  of  inertia  of 275 

Permanent  set 16 

Physical  properties 2 

Piers 106 

Pipe,  stresses  in  a  thin 25 

Plasticity 2 

Plate  girders 212 

Points,  of  inflection 157 

hinging,  for  continuous  beams 185 

Poisson's  ratio ig 

Polar  moment  of  inertia 273 


INDEX  305 

PAGE 

Pole 121 

Pole  distance 121 

Prisms,  eccentric  loads  on  short 107 

Properties,  mechanical  and  physical 2 

Radius  of  curvature  of  beams 130 

Radius  of  gyration 273 

Rankine's  formula 231 

Ray  polygon 121 

Rays 121 

Reduction  of  area 19 

Relations  between  the  five  curves 137 

algebraic 189 

Relative  strength  and  stiffness  of  beams 160 

Repeated  stresses 255 

Resilience 18,  256 

elastic 256 

modulus  of 256 

of  a  bar  under  direct  stress 258 

of  a  beam 258 

Resisting  moment 86 

Resisting  shear 84 

Rest,  the  effect  of 260 

Riveted  joints 32 

computation  of  unit-stresses  developed  in 34 

efficiency  of 38 

kinds  of 32 

methods  of  failure  of 33 

Rupture,  modulus  of 97 

Second  integrated  curve,  the  first  method  of  obtaining  the 117 

the  second  method  of  obtaining  the 119 

Section,  cross 10 

Section  modulus 90 

Shafts,  round,  stress  and  deformation 246 

other  shapes  of  cross  section  of 249 

power  transmitted  by .  . 250 

stiffness  of 248 

Shear  formula,  the •. 84 

values  of  k  in  the 85 

Shear 13 

double 35 


306  INDEX 

PAGE 

Shear,  horizontal,  in  beams 208 

oblique 13 

resisting 84 

single 35 

vertical 52 

vertical,  sign  and  unit  of 53 

vertical,  values  of 53 

Shearing  modulus  of  elasticity 18 

Shearing  stress,  the  maximum 209 

combined,  and  tensile  or  compressive  stress 214 

Shell,  stresses  in  a  thin 25 

Shortening 9 

Simple  beam 47,  65,  73,  148,  150 

Slenderness  ratio 222 

Slope  curve 134 

Slope  of  the  neutral  surface 132 

Slope,  the  rate  of  change  of 135 

Spheres,  stresses  in  thin 28 

Steel 5 

Stiffness 2 

Stiffness  of  beams,  relative 160 

Stiffness  of  shafts 248 

Straight-line  formula,  the 225 

Strength  of  beams,  relative 73 

Strength,  rupturing 1 1 

ultimate  compressive 12 

ultimate  shearing 13 

ultimate  tensile n 

Stress 9 

axial 10 

combined  shearing  and  tensile,  or  compressive 216 

fiber 90 

maximum,  diagrams 97 

maximum  horizontal  and  vertical  shearing  unit,  at  a  point 209 

relation  between  the  maximum  stress  and  the  maximum  deflection  164 

Stress-deformation  diagrams 15 

Stresses,  compressive 12 

due  to  change  in  temperature 27 

fiber,  distribution  of : 87 

flexural 47 

in  a  hoop 26 

internal 51 


INDEX  307 

PAGE 

Stresses,  in  thin  cylinders 25 

in  thin  spheres 28 

maximum 209 

repeated 255 

shearing 13 

tensile 10 

total  horizontal  compressive  and  tensile 91 

used  in  design 20 

working 20 

String  polygon 121 

Strings 121 

Struts,  columns  and 221 

Suddenly  applied  loads 261 

Surface,  neutral 87,  88 

Symmetry,  axis  of 268 

Temperature,  stresses  due  to  change  in 27 

Tension 10 

combined  flexure  and,  or  compression 214 

Theorem  of  three  moments 181.  205 

Timber ' 4 

Torsion 246 

the,  formula 247 

Toughness 3 

Triangle,  centroid  of  a 268 

moment  of  inertia  of  a 279 

Twisting,  combined,  and  bending 251 

Uniform  load 49 

Unit-stress 9 

Units  for  the  five  curves 137 

Varying  load 49 

Vector  polygon 121 

Vertical  shear 52 

diagrams 54 

diagrams  for  cantilever  and  simple  beams 65 

maximum,  the 64,  65 

rate  of  change  of 57 

relation  between,  and  the  bending  moment 62 

sign  and  unit  of 53 

values  of 53 

values  of  the  maximum 64,  65 


308  INDEX 

PAGE 

Walls 106 

Wrought  iron 5 

Yield  point 16 

Young's  modulus " 17 

Zero  bending  moment,  the  section  of 196 

Zero  deflection,  the  section  of 196 

Zero  slope,  the  section  of 196 

Zero  vertical  shear,  the  section  of 195 


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